Partial Differentiation

We sometimes come across quantities that are functions of more than one independent variable. Thus, we may find a case where $y$ depends on two other variable quantities, one of which we will call $u$ and the other $v$ . In symbols $y = f (u, v) .$ Take the simplest concrete case.

Let $y = u \times v .$ What are we to do? If we were to treat $v$ as a constant, and differentiate with respect to $u$ , we should get $d y_{v} = v d u;$ or if we treat $u$ as a constant, and differentiate with respect to $v$ , we should have: $d y_{u} = u d v .$

The little letters here put as subscripts are to show which quantity has been taken as constant in the operation.

Another way of indicating that the differentiation has been performed only partially, that is, has been performed only with respect to one of the independent variables, is to write the derivatives with a “curved dee” $\partial$ , instead of the regular letter $d$ . In this way \begin{align} \strut \frac{\partial y}{\partial u} &= v, \\ \frac{\partial y}{\partial v} &= u. \end{align}

If we put in these values for $v$ and $u$ respectively, we shall have \left. \begin{align} \strut dy_v &= \frac{\partial y}{\partial u}\, du, \\ dy_u &= \frac{\partial y}{\partial v}\, dv, \end{align} \right\} \quad\text{which are }\textbf{partial differentials}.

But, if you think of it, you will observe that the total variation of $y$ depends on both these things at the same time. That is to say, if both are varying, the real $d y$ ought to be written $d y = \frac{\partial y}{\partial u} d u + \frac{\partial y}{\partial v} d v;$ and this is called a total differential.

Example 16.1. Find the partial derivatives of the expression $w = 2 a x^{2} + 3 b x y + 4 c y^{3}$ .

Solution. The answers are: \left. \begin{align} \frac{\partial w}{\partial x} &= 4ax + 3by. \\ \frac{\partial w}{\partial y} &= 3bx + 12cy^2. \end{align} \right\}

The first is obtained by supposing $y$ constant, the second is obtained by supposing $x$ constant; then $d w = (4 a x + 3 b y) d x + (3 b x + 12 c y^{2}) d y .$

Example 16.2. Let $z = x^{y}$ . Then, treating first $y$ and then $x$ as constant, we get in the usual way \left. \begin{align} \dfrac{\partial z}{\partial x} &= yx^{y-1}, \\ \dfrac{\partial z}{\partial y} &= x^y \times \ln x, \end{align}\right\} so that $d z = y x^{y - 1} d x + x^{y} \ln x d y$ .

Example 16.3. A cone having height $h$ and radius of base $r$ has volume $V = \frac{1}{3} π r^{2} h$ . If its height remains constant, while $r$ changes, the ratio of change of volume, with respect to radius, is different from ratio of change of volume with respect to height which would occur if the height were varied and the radius kept constant, for \left. \begin{align} \frac{\partial V}{\partial r} &= \dfrac{2\pi}{3} rh, \\ \frac{\partial V}{\partial h} &= \dfrac{\pi}{3} r^2. \end{align}\right\}

The variation when both the radius and the height change is given by $d V = \frac{2 π}{3} r h d V + \frac{π}{3} r^{2} d h$ .

Example 16.4. In the following example $F$ and $f$ denote two arbitrary functions of any form whatsoever. For example, they may be sine-functions, or exponentials, or mere algebraic functions of the two independent variables, $t$ and $x$ . This being understood, let us take the expression $y = F (x + a t) + f (x - a t),$ or $y = F (w) + f (v);$ where $w = x + a t$ , and $v = x - a t$ .
Then \begin{align} \frac{\partial y}{\partial x} &= \frac{d F(w)}{d w}\cdot\frac{\partial w}{\partial x} + \frac{d f(v)}{d v}\cdot\frac{\partial v}{\partial x} \\ &= F^\prime(w)\cdot1 + f^\prime(v)\cdot1 \end{align} (where the figure $1$ is simply the coefficient of $x$ in $w$ and $v$ );
and \begin{align} \frac{\partial^2 y}{\partial x^2}= F^{\prime\prime}(w) + f^{\prime\prime}(v). \end{align} Also \begin{align} \frac{\partial y}{\partial t} &= \frac{d F(w)}{d w}\cdot\frac{\partial w}{\partial t} + \frac{d f(v)}{d v}\cdot\frac{\partial v}{\partial t} \\ &= F^\prime(w)\cdot a - f^\prime(v) a; \end{align} and \begin{align} \frac{\partial^2 y}{\partial t^2} &= F''(w)a^2 + f''(v)a^2; \end{align} whence \begin{align} \frac{\partial^2 y}{\partial t^2} &= a^2\, \frac{\partial^2 y}{\partial x^2}. \end{align}

This differential equation is of immense importance in mathematical physics.

Maxima and Minima of Functions of two Independent Variables

Let us take up again the following exercise from the chapter on Maxima and Minima:

Example 16.5. A piece of string $30$ inches long has its two ends joined together and is stretched by $3$ pegs so as to form a triangle. What is the largest triangular area that can be enclosed by the string?

Solution. Let $x$ and $y$ be the length of two of the portions of the string. The third is $30 - (x + y)$ , and the area of the triangle is $A = \sqrt{s (s - x) (s - y) (s - 30 + x + y)}$ , where $s$ is the half perimeter, $15$ , so that $A = \sqrt{15 P}$ , where \begin{align} P &= (15-x)(15-y)(x+y-15) \\ &= xy^2 + x^2y - 15x^2 - 15y^2 - 45xy + 450x + 450y - 3375. \end{align}

Clearly $A$ is maximum when $P$ is maximum. $d P = \frac{\partial P}{\partial x} d x + \frac{\partial P}{\partial y} d y .$ For a maximum (clearly it will not be a minimum in this case), one must have simultaneously

$\frac{\partial P}{\partial x} = 0 and \frac{\partial P}{\partial y} = 0;$

that is, \begin{align} 2xy - 30x + y^2 - 45y + 450 &= 0, \\ 2xy - 30y + x^2 - 45x + 450 &= 0. \end{align}

An immediate solution is $x = y$ .

If we now introduce this condition in the value of $P$ , we find $P = (15 - x)^{2} (2 x - 15) = 2 x^{3} - 75 x^{2} + 900 x - 3375.$ Now $P$ is a function of $x$ alone. For maximum or minimum, $\frac{d P}{d x} = 6 x^{2} - 150 x + 900 = 0$ , which gives $x = 15$ or $x = 10$ .

Clearly $x = 15$ gives minimum area; $x = 10$ gives the maximum, for $\frac{d^{2} P}{d x^{2}} = 12 x - 150$ , which is $+ 30$ for $x = 15$ and $- 30$ for $x = 10$ (see the Second Derivative Test).

Example 16.6. Find the dimensions of an ordinary railway coal truck with rectangular ends, so that, for a given volume $V$ the area of sides and floor together is as small as possible.

Solution. The truck is a rectangular box open at the top. Let $x$ be the length and $y$ be the width; then the depth is $\frac{V}{x y}$ . The surface area is $S = x y + \frac{2 V}{x} + \frac{2 V}{y}$ . \begin{align} dS &= \frac{\partial S}{\partial x}\, dx + \frac{\partial S}{\partial y}\, dy\\ &= \left(y - \frac{2V}{x^2}\right) dx + \left(x - \frac{2V}{y^2}\right) dy. \end{align} For minimum (clearly it won’t be a maximum here), $y - \frac{2 V}{x^{2}} = 0, x - \frac{2 V}{y^{2}} = 0.$

Here also, an immediate solution is $x = y$ , so that $S = x^{2} + \frac{4 V}{x}$ , $\frac{d S}{d x} = 2 x - \frac{4 V}{x^{2}} = 0$ for minimum, and $x = \sqrt[3]{2 V} .$

Exercises

Exercise 16.1. Differentiate the expression $\frac{x^{3}}{3} - 2 x^{3} y - 2 y^{2} x + \frac{y}{3}$ with respect to $x$ alone, and with respect to $y$ alone.

Answer

$x^{3} - 6 x^{2} y - 2 y^{2}; \frac{1}{3} - 2 x^{3} - 4 x y$ .

Solution

\begin{align} & z=\frac{x^{3}}{3}-2 x^{3} y-2 y^{2} x+\frac{y}{3} \\ & \frac{\partial z}{\partial x}=x^{2}-6 x^{2} y-2 y^{2} \\ & \frac{\partial z}{\partial y}=-2 x^{3}-4 y x+\frac{1}{3} \end{align}

Exercise 16.2. Find the partial derivatives with respect to $x$ , $y$ and $z$ , of the expression $x^{2} y z + x y^{2} z + x y z^{2} + x^{2} y^{2} z^{2} .$

Answer

$2 x y z + y^{2} z + z^{2} y + 2 x y^{2} z^{2}$ ;
$2 x y z + x^{2} z + x z^{2} + 2 x^{2} y z^{2}$ ;
$2 x y z + x^{2} y + x y^{2} + 2 x^{2} y^{2} z$ .

Solution

Let $u = x^{2} y z + x y^{2} z + x y z^{2} + x^{2} y^{2} z^{2} .$ Then \begin{align} \frac{\partial u}{\partial x}= & 2 x y z+y^{2} z+y z^{2}+2 x y^{2} z^{2} \\ \frac{\partial u}{\partial y}= & x^{2} z+2 x y z+x z^{2}+2 x^{2} y z \\ \frac{\partial u}{\partial z}= & x^{2} y+x y^{2}+2 x y z+2 x^{2} y^{2} z \end{align}

Exercise 16.3. Let $r^{2} = (x - a)^{2} + (y - b)^{2} + (z - c)^{2}$ .

Find the value of $\frac{\partial r}{\partial x} + \frac{\partial r}{\partial y} + \frac{\partial r}{\partial z}$ . Also find the value of $\frac{\partial^{2} r}{\partial x^{2}} + \frac{\partial^{2} r}{\partial y^{2}} + \frac{\partial^{2} r}{\partial z^{2}}$ .

Answer

$\frac{1}{r} {(x - a) + (y - b) + (z - c)} = \frac{(x + y + z) - (a + b + c)}{r}$ ; $\frac{3}{r}$ .

Solution

\begin{align} & r^{2}=(x-a)^{2}+(y-b)^{2}+(z-c)^{2} \\ & \frac{\partial\left(r^{2}\right)}{\partial x}=2 r \frac{\partial r}{\partial x}=2(x-a) \\ \Rightarrow\qquad & \frac{\partial r}{\partial x}=\frac{x-a}{r} \end{align} Similarly

$2 r \frac{\partial r}{\partial y} = 2 (y - b) \Rightarrow \frac{\partial r}{\partial y} = \frac{y - b}{r}$ $2 r \frac{\partial r}{\partial z} = 2 (z - c) \Rightarrow \frac{\partial r}{\partial z} = \frac{z - c}{r} .$ Therefore, \begin{align} \frac{\partial r}{\partial x}+\frac{\partial r}{\partial y}+\frac{\partial r}{\partial z} & =\frac{x-a+y-b+z-c}{r} \\ & =\frac{x+y+z}{r}-\frac{a+b+c}{r} \end{align}

Using the Quotient Rule

\begin{align} \frac{\partial^{2} r}{\partial x^{2}} &= \frac{\partial\left(\dfrac{\partial r}{\partial x}\right)}{\partial x}\\ &=\frac{\partial\left(\dfrac{x-a}{r}\right)}{\partial x}\\ &=\frac{r-\dfrac{\partial r}{\partial x}(x-a)}{r^{2}} &&\text{(Quotient Rule)}\\ & =\frac{r-\frac{(x-a)^{2}}{r}}{r^{2}} \\ & =\frac{r^{2}-(x-a)^{2}}{r^{3}} \end{align} Similarly $\frac{\partial^{2} r}{\partial y^{2}} = \frac{r^{2} - (y - b)^{2}}{r^{3}}$ and $\frac{\partial^{2} r}{\partial z^{2}} = \frac{r^{2} - (z - c)^{2}}{r^{3}}$

Hence

\begin{align} \frac{\partial^{2} r}{\partial x^{2}}+\frac{\partial^{2} r}{\partial y^{2}}+\frac{\partial^{2} r}{\partial z^{2}} & =\frac{r^{2}-(x-a)^{2}}{r^{3}}+\frac{r^{2}-(y-b)^{2}}{r^{3}}+\frac{r^{2}-(z-c)^{2}}{r^{3}} \\ & =\frac{3 r^{2}-\left\{(x-a)^{2}+(y-b)^{2}+(z-c)^{2}\right\}}{r^{3}} \\ & =\frac{3 r^{2}-r^{2}}{r^{3}} \\ & =\frac{2}{r} . \end{align}

Exercise 16.4. Find the total differential of $y = u^{v}$ .

Answer

$d y = v u^{v - 1} d u + u^{v} \ln u d v$ .

Solution

$\begin{matrix} y = u^{v} \\ \frac{\partial y}{\partial u} = v u^{v - 1} \frac{\partial y}{\partial v} = u^{v} \cdot \ln u \end{matrix}$ The total differential of $y$ is \begin{align} & d y=\frac{\partial y}{\partial u} d u+\frac{\partial y}{\partial v} d v \\ & =v u^{v-1} d u+u^{v} \cdot \ln u d v \end{align}

Exercise 16.5. Find the total differential of $y = u^{3} \sin v$ ; of $y = (\sin x)^{u}$ ; and of $y = \frac{\ln u}{v}$ .

Answer

$d y = 3 \sin v u^{2} d u + u^{3} \cos v d v$ ,
$d y = u {(\sin x)}^{u - 1} \cos x d x + (\sin x)^{u} \ln \sin x d u$ ,
$d y = \frac{1}{v} \frac{1}{u} d u - \ln u \frac{1}{v^{2}} d v$ .

Solution

(a) $y = u^{3} \sin v$

\begin{align} & \frac{\partial y}{\partial u}=3 u^{2} \sin v \quad \frac{\partial y}{\partial v}=u^{3} \cos v \\ & d y=\frac{\partial y}{\partial u} d u+\frac{\partial y}{\partial v} d v \end{align}

(b) $y = (\sin x)^{u}$

\begin{align} & \frac{\partial y}{\partial x}=u \cos x(\sin x)^{u-1} \\ & \frac{\partial y}{\partial u}=(\sin x)^{u} \cdot \ln (\sin x) \\ & d y=\frac{\partial y}{\partial x} d x+\frac{\partial y}{\partial u} d u \\ & d y=u \cos x(\sin x)^{u-1} d x+(\sin x)^{u} \cdot \ln (\sin x) d u \end{align}

(c)

\begin{align} y & =\frac{\ln u}{v} \\ \frac{\partial y}{\partial u} & =\frac{1}{v} \frac{1}{u} \\ \frac{\partial y}{\partial v} & =\frac{\partial\left(\ln u \cdot v^{-1}\right)}{\partial v} \\ & =-\ln u \cdot v^{-2} \\ & =-\frac{\ln u}{v^{2}} \\ d y & =\frac{\partial y}{\partial u} d u+\frac{\partial y}{\partial v} d v \\ & =\frac{1}{u v} d u-\frac{\ln u}{v^{2}} d v \end{align}

Exercise 16.6. Verify that the sum of three quantities $x$ , $y$ , $z$ , whose product is a constant $k$ , is maximum when these three quantities are equal.

Solution

We want to maximize

$S = x + y + z$

provided that $x y z = k$ .

We find $z$ from the constraint $x y z = k$ and then substitute it in $S$ : $x y z = k \Rightarrow z = \frac{k}{x y}$ $S = x + y + z = x + y + \frac{k}{x y}$

The maximum occurs when $\frac{\partial S}{\partial x} = \frac{\partial S}{\partial y} = 0$ \left\{ \begin{aligned} \frac { \partial S } { \partial x } &= 1 - \frac { k } { x ^ { 2 } y } = 0 \\ \frac { \partial S } { \partial y } &= 1 - \frac { k } { x y ^ { 2 } } = 0 \end{aligned} \right. \Rightarrow \left\{ \begin{aligned} &x^{2} y=k \\ &x y^{2}=k \end{aligned}\right. If we divide $x^{2} y = k$ by $x y^{2} = k$ , we get $\frac{x^{2} y}{x y^{2}} = \frac{k}{k}$ Simplifying the left side, we get: $\frac{x}{y} = 1 \Rightarrow x = y$ $x^{2} y = k \Rightarrow x^{3} = k \Rightarrow x = \sqrt[3]{k}$ Therefore $y = \sqrt[3]{k}$ and $z = \frac{k}{x y} = \frac{k}{\sqrt[3]{k} \sqrt[3]{k}} = \sqrt[3]{k} .$

Exercise 16.7. Find the maximum or minimum of the function $u = x + 2 x y + y .$

Answer

Minimum for $x = y = - \frac{1}{2}$ .

Solution

\begin{align} &\frac{\partial u}{\partial x}=1+2 y\\ &\frac{\partial u}{\partial y}=1+2 x \end{align}

$u$ has a maximum or minimum where $\frac{\partial u}{\partial x} = \frac{\partial u}{\partial y} = 0$ :

\begin{align} & \frac{\partial u}{\partial x}=1+2 y=0 \Rightarrow y=-\frac{1}{2} \\ & \frac{\partial u}{\partial y}=1+2 x=0 \Rightarrow x=-\frac{1}{2} \end{align}

By examining nearby points (such as $x = - 0.4, y = - 0.4$ ), we can say that $u$ is a minimum when $x = y = - \frac{1}{2}$ .

Exercise 16.8. The post-office regulations state that no parcel is to be of such a size that its length plus its girth exceeds $6$ feet. What is the greatest volume that can be sent by post (a) in the case of a package of rectangular cross section; (b) in the case of a package of circular cross section?

Answer

(a) Length $2$ feet, width = depth = $1$ foot, vol. = $2$ cubic feet.
(b) Radius = $\frac{2}{π}$ feet = $7.46$ in., length = $2$ feet, vol. = $2.546$ .

Solution

(a) Let \begin{align} & L=\text { length of parcel } \\ & W=\text { Width } \\ & H=\text { Height } \end{align} Then $girth = L + 2 W + 2 H$

We want to maximize $V = L W H$ provided that $girth = L + 2 W + 2 H = 6$ or $L = 6 - 2 W - 2 H .$

Using this constraint, we can write $V$ as \begin{align} V & =(6-2 W-2 H) W H \\ & =6 W H-2 W^{2} H-2 \end{align}

The maximum occurs where $\frac{\partial V}{\partial W} = 0$ and $\frac{\partial V}{\partial H} = 0$ : \left\{\begin{align} & \frac{\partial V}{\partial W}=6 H-4 W H-2 H^{2}=0 \\ & \frac{\partial V}{\partial H}=6 W-W^{2}-4 W H=0 \end{align}\right. Subtracting the second equation from the first one $6 (H - W) - 2 (H^{2} - W^{2}) = 0$ or \begin{align} & 6(H-W)-2(H-W)(H+W)=0 \\ & (H-W)[6-2 H-2 W]=0 \end{align}

The expression between brackets is $L$ and since $L \neq 0$ , the only solution is $H - W = 0$ or $W = H$ .

Substituting $W = H$ in $\frac{\partial V}{\partial x} = 0$ yields $6 H - 4 H^{2} - 2 H^{2} = 6 H - 6 H^{2} = 0$ or $6 H (1 - H) = 0$ Since $H \neq 0$ $\Rightarrow H = 1 \Rightarrow W = H = 1$ Therefore, \begin{align} & L=6-2 H-2 W \\ & L=2 \end{align} [Alternatively we could substitute $W = H$ in $\frac{\partial V}{\partial y} = 0$ .]

In this case $(H = W = 1$ and $L = 2), V = 2 {ft}^{3}$ .

(b) $r =$ radius of the cross section

$\begin{matrix} girth = L + 2 π r \\ L + 2 π r = 6 \Rightarrow L = 6 - 2 π r \end{matrix}$

We want to maximize $V = π r^{2} L .$

Since $L = 6 - 2 π r$ : \begin{align} V& =\pi r^{2}(6-2 \pi r) \\ & =6 \pi r^{2}-2 \pi^{2} r^{3} \end{align}

Now $V$ is a function of $r$ alone:

\begin{align} \frac{d V}{d r} & =12 \pi r-6 \pi^{2} r^{2} \\ & =6 \pi r(2-\pi r) \end{align} $\frac{d V}{d r} = 0 \Leftrightarrow r = 0 or r = \frac{2}{π}$

Since $r \neq 0$ , the only solution is $r = \frac{2}{π}$ .

When $r = \frac{2}{π}$ :

$L = 6 - 2 π \times \frac{2}{π} = 2 f t$

and the maximum volume is obtained when $r = \frac{2}{π}$ and $L = 2$

$V = π {(\frac{2}{π})}^{2} \cdot 2 = \frac{8}{π} \approx 2.546 {ft}^{3} .$

Exercise 16.9. Divide $π$ into $3$ parts such that the continued product of their sines may be a maximum or minimum.

Answer

All three parts equal; the product is maximum.

Solution

We want to maximize

$u = \sin x \sin y \sin z$ provided that $x + y + z = π, (x, y, z \geq 0)$ . Also none of them can be zero because $u$ will be zero if $x = 0$ or $y = 0$ or $z = 0$ of $u$ occurs if $x = y = z = \frac{π}{3}$ . By symmetry we can imagine the maximum of $u$ occurs if $x = y = z = \frac{π}{3}$

But here we want to use the calculus: $x + y + z = π \Rightarrow z = π - x - y$ and thus $u = \sin x \sin y \sin (π - x - y)$ Recall that $\sin (π - θ) = \sin θ$ . Hence

$u = \sin x \sin y \sin (x + y) .$

To maximize $u$ :

\left\{ \begin{align} \frac{\partial u}{\partial x}&=\cos x \sin y \sin (x+y)+\sin x \sin y \cos (x+y)=0 &&\text{(A)} \\ \frac{\partial u}{\partial y}&=\sin x\cos y\sin(x+y)+\sin x\sin y \cos(x+y)=0 &&\text{(B)} \end{align} \right. Subtracting (A) from (B), we obtain $[\sin x \cos y - \cos x \sin y] \sin (x + y) = 0$ Since $\sin (A - B) = \sin A \cos B - \cos A \sin B$ , we can write the expression between brackets as $\sin (x - y)$ . Hence

$\sin (x - y) \sin (x + y) = 0$

$\Rightarrow x - y = 0, or x - y = π, or x + y = 0, or x + y = π .$

since $0 < x, y < π$ , only $x - y = 0$ is acceptable:

$x - y = 0 \Rightarrow x = y .$

Using $x = y$ in (A), we get

$\cos x \sin x \sin (2 x) + \sin^{2} x \cos (2 x) = 0$

Note $\sin 2 x = 2 \sin x \cos x$ :

$\begin{matrix} 2 \cos^{2} x \sin^{2} x + \sin^{2} x \cos 2 x = 0 \\ \sin^{2} x (2 \cos^{2} x + \cos 2 x) = 0 \\ \cos 2 x = 2 \cos^{2} x - 1 : \\ \sin^{2} x (4 \cos^{2} x - 1) = 0 \end{matrix}$

Since $0 < x < π, \sin x \neq 0$ ; therefore

$\sin^{2} x (4 \cos^{2} x - 1) = 0 \Leftrightarrow \cos^{2} x = \frac{1}{4}$ or $\cos x = \frac{1}{2} \Rightarrow x = \frac{π}{3}$

This means $u$ is a maximum when $x = y = z = \frac{π}{3}$ and the maximum value of $u$ is

$\sin^{3} (\frac{π}{3}) = {(\frac{\sqrt{3}}{2})}^{3} = \frac{3^{\frac{3}{2}}}{8} \approx 0.65$

Exercise 16.10. Find the maximum or minimum of $u = \frac{e^{x + y}}{x y}$ .

Answer

Minimum for $x = y = 1$ .

Solution

$u = \frac{e^{x + y}}{x y} = \frac{e^{x}}{x} \cdot \frac{e^{y}}{y}$

\begin{align} \frac{\partial u}{\partial x}&=-\frac{e^{x}}{x^{2}} \cdot \frac{e^{y}}{y}+\frac{e^{x}}{x} \cdot \frac{e^{y}}{y}\\ &=\frac{e^{x}}{x} \cdot \frac{e^{y}}{y}\left(-\frac{1}{x}+1\right)=0 \\ \frac{\partial u}{\partial y}&=\frac{e^{x}}{x}\left(-\frac{e^{y}}{y^{2}}\right)+\frac{e^{x}}{x} \cdot \frac{e^{y}}{y}\\ &=\frac{e^{x}}{x} \cdot \frac{e^{y}}{y}\left(-\frac{1}{y}+1\right)=0 \end{align}

Since $\frac{e^{x + y}}{x y} \neq 0$ , we must have

$- \frac{1}{x} + 1 = 0 and - \frac{1}{y} + 1 = 0$ or $x = 1 and y = 1$

When $x = y = 1, u = e^{2} \approx 7.39$ .

Let’s examine some nearby points:

\begin{align} & x=y=1.1, \quad u=\frac{e^{2.2}}{1.1^{2}} \approx 7.46 \\ & x=y=0.9, \quad u=\frac{e^{1.8}}{0.9^{2}} \approx 7.47 \\ & x=1.1, \quad y=0.9, \quad u=\frac{e^{2}}{1.1 \times 0.9} \approx 7.46 \end{align}

Therefore, $u$ has a minimum of $e^{2}$ .

Exercise 16.11. Find maximum and minimum of $u = y + 2 x - 2 \ln y - \ln x .$

Answer

Min.: $x = \frac{1}{2}$ and $y = 2$ .

Solution

$u = y + 2 x - 2 \ln y - \ln x$ \begin{align} & \frac{\partial u}{\partial x}=2-\frac{1}{x}=0 \Rightarrow x=\frac{1}{2} \\ & \frac{\partial u}{\partial y}=1-\frac{2}{y}=0 \Rightarrow y=2 \end{align}

When $x = 0.5$ and $y = 2$ , $u = 3 - \ln 2 \approx 2.31 .$

We can examine a couple of points near $x = \frac{1}{2}, y = 2$

When $x = 0.4, y = 1.9, u \approx 2.33$

When $x = 0.4, y = 2.1, u \approx 2.33$

When $x = 0.6, y = 1.9, u \approx 2.33$

When $x = 0.6, y = 2.1, u \approx 2.33$

Hence, $u$ is a minimum when $x = \frac{1}{2}$ and $y = 2$ .

Exercise 16.12. A telpherage bucket of given capacity has the shape of a horizontal isosceles triangular prism with the apex underneath, and the opposite face open. Find its dimensions in order that the least amount of iron sheet may be used in its construction.

Answer

Angle at apex $= 90^{\circ}$ ; equal sides = length = $\sqrt[3]{2 V}$ .

Solution

$Area of the triangle = A_{1} = \frac{1}{2} x^{2} \sin θ$

$Volume = V = A_{1} L = \frac{1}{2} x^{2} \sin θ L$

\begin{align} \text{Area of the bucket } =A&=2 A_{1}+2 x L\\ &=x^{2} \sin \theta+2 x L \end{align}

$V = A_{1} L = \frac{1}{2} x^{2} \sin θ L \Rightarrow L = \frac{2 V}{x^{2} \sin θ}$

$A = x^{2} \sin θ + \frac{4 V}{x \sin θ}$

To minimize, we set $\frac{\partial A}{\partial x} = 0$ and $\frac{\partial A}{\partial θ} = 0$

\left\{\begin{align} &\frac{\partial A}{\partial x}=2 x \sin \theta-\frac{4 V}{x^{2} \sin \theta}=0 \\[9pt] &\frac{\partial A}{\partial y}=x^{2} \cos \theta-\frac{4 V}{x \sin ^{2} \theta} \cos \theta=0 \end{align}\right.

\left\{\begin{align} &x \sin \theta-\frac{2 v}{x^{2} \sin \theta}=0 \\[9pt] &\cos \theta\left[x^{2}-\frac{4 V}{x \sin ^{2} \theta}\right]=0 \end{align}\right.

Since $\sin θ \neq 0$ and $x \neq 0$ , we multiply both sides of the first equation by $x$ and divide them by $\sin θ$ :

\left\{\begin{align} &x^{2}-\frac{2 V}{x \sin ^{2} \theta}=0 \\ &\cos \theta\left[x^{2}-\frac{4 V}{x \sin ^{2} \theta}\right]=\cos \theta\left[\underbrace{x^{2}-\frac{2 V}{x \sin ^{2} \theta}}_{0}-\frac{2 V}{x \sin ^{2} \theta}\right]=0 \end{align}\right.

So the second equation can be written as

$\cos θ \frac{2 V}{x \sin^{2} θ} = 0$

Since $V \neq 0$ , we must have $\cos θ = 0$ or

$θ = \frac{π}{2} (or 90^{\circ})$

Since $\sin \frac{π}{2} = 1$ , from the first equation we obtain $x^{2} - \frac{2 V}{x} = 0 \Rightarrow x = \sqrt[3]{2 V}$ and because $L = \frac{2 V}{x^{2} \sin θ}$

$L = \frac{2 V}{(2 V)^{\frac{2}{3}}} = (2 V)^{\frac{1}{3}} = \sqrt[3]{2 V}$

Partial Differentiation

Maxima and Minima of Functions of two Independent Variables

Exercises

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