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PARTIAL DIFFERENTIATION

179

triangle is $A={\sqrt {s(s-x)(s-y)(s-30+x+y)}}$ , where $s$ is the half perimeter, $15$ , so that $A={\sqrt {15P}}$ , where

${\begin{aligned}P&=(15-x)(15-y)(x+y-15)\\&=xy^{2}+x^{2}y-15x^{2}-15y^{2}-45xy+450x+450y-3375.\end{aligned}}$

Clearly $A$ is maximum when $P$ is maximum.

$dP={\dfrac {\partial P}{\partial x}}\,dx+{\dfrac {\partial P}{\partial y}}\,dy$ .

For a maximum (clearly it will not be a minimum in this case), one must have simultaneously

${\dfrac {\partial P}{\partial x}}=0\quad {\text{and}}\quad {\dfrac {\partial P}{\partial y}}=0$ ;

that is,

$\left.{\begin{aligned}2xy-30x+y^{2}-45y+450&=0,\\2xy-30y+x^{2}-45x+450&=0.\end{aligned}}\right\}$

An immediate solution is $x=y$ .

If we now introduce this condition in the value of $P$ , we find

$P=(15-x)^{2}(2x-15)=2x^{3}-75x^{2}+900x-3375$ .

For maximum or minimum, ${\dfrac {dP}{dx}}=6x^{2}-150x+900=0$ , which gives $x=15$ or $x=10$ .

Clearly $x=15$ gives minimum area; $x=10$ gives the maximum, for ${\dfrac {d^{2}P}{dx^{2}}}=12x-150$ , which is $+30$ for $x=15$ and $-30$ for $x=10$ .

Example (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.

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