Answer to Question #271248 in Calculus for ozair

Question

You plan to make a simple, open topped box from a piece of sheet metal by cutting a square – of equal size – from each corner and folding up the sides as shown in the diagram: If 𝑙 = 200𝑚𝑚 and 𝑤 = 150𝑚𝑚 calculate: a) The value of x which will give the maximum volume b) The maximum volume of the box c) Comment of the value obtained in part b.

Answer 1

a) Let $x=$ the length of side of a cutted square

Then the volume of a box will be

$$V=x(w-2x)(l-2x), 0<x<w/2, 0<x<l/2$$

Given $l=200\ mm, w=150\ mm$

$$V=V(x)=x(150-2x)(200-2x), 0<x<75$$

$$V(x)=4(75x-x^2)(100-x)$$

$$=4(7500x-175x^2+x^3)$$

Differentiate with respect to $x$

$$V'(x)=4(7500-350x+3x^2)$$

Find the critical number(s)

$$V'(x)=0=>4(7500-350x+3x^2)=0$$

$$3x^2-350x+7500=0$$

$$D=(-30)^2-4(3)(7500)=32500$$

$$x=\dfrac{350\pm\sqrt{32500}}{2(3)}=\dfrac{175\pm25\sqrt{13}}{3}$$

Critical numbers: $\dfrac{175-25\sqrt{13}}{3}, \dfrac{175+25\sqrt{13}}{3}$

If $x<\dfrac{175-25\sqrt{13}}{3}, V'(x)>0 , V(x)$ increases.

If $\dfrac{175-25\sqrt{13}}{3}<x<\dfrac{175+25\sqrt{13}}{3}, V'(x)<0 ,$

$V(x)$ decreases.

If $x>\dfrac{175+25\sqrt{13}}{3}, V'(x)>0 , V(x)$ increases.

Since $0<x<75,$ we consider

If $0<x<\dfrac{175-25\sqrt{13}}{3}, V'(x)>0 , V(x)$ increases.

If $\dfrac{175-25\sqrt{13}}{3}<x<75, V'(x)<0 ,$ $V(x)$ decreases.

The volume has the absolute maximum on $(0, 75)$ at $x=\dfrac{175-25\sqrt{13}}{3}.$

$x=\dfrac{175-25\sqrt{13}}{3}.$

b)

$$V_{max}=4\cdot\dfrac{175-25\sqrt{13}}{3}\cdot(75-\dfrac{175-25\sqrt{13}}{3})$$

$$\cdot(100-\dfrac{175-25\sqrt{13}}{3})$$

$$=\dfrac{62500}{27}(7-\sqrt{13})(2+\sqrt{13})(5+\sqrt{13})$$

$$=\dfrac{125000}{27}(35+13\sqrt{13})\ ({mm}^3)$$

c) Since the number $\dfrac{175-25\sqrt{13}}{3}$ is irrational we take $x\approx28.3\ mm.$

$$V_{max}=4(28.3)(75-28.3)(100-28.3)=379037.748$$

$$=379037.748\ ({mm}^3)\approx379\ cm^3$$