Page:Calculus Made Easy.pdf/199

 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

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

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

that is, {{c|$$\left. \begin{aligned} 2xy - 30x + y^2 - 45y + 450 &= 0, \\ 2xy - 30y + x^2 - 45x + 450 &= 0. \end{aligned}\right\}$$}} An immediate solution is $$x=y$$.

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

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.