Answer to Question #259058 in Calculus for Ale

Question

From a thin piece of cardboard 10 in. by 10 in., square corners are cut out so that the sides can be folded up to make a box. What dimensions will yield a box of maximum volume? What is the maximum volume? Round to the nearest tenth, if necessary.

A. 6.7 in. x 6.7 in. x 3.3 in.; 148.1 in³

B. 3.3 in. x 3.3 in. x 3.3 in.; 37 in³

C. 5 in. x 5 in. x 2.5 in.; 62.5 in³

D. 6.7 in. x 6.7 in. x 1.7 in.; 74.1 in³

Answer 1

Let $x=$ the side of cutted square corner. Then a volume of a box will e

$$V=V(x)=(10-2x)^2x, \ 0\leq x\leq5$$

Find the first derivative with respect to $x$

$$V'(x)=((10-2x)^2x)'$$

$$=(10-2x)^2+2(10-2x)(-2)x$$

$$=(10-2x)(10-2x-4x)=4(5-x)(5-3x)$$

Find the critical number(s)

$$V'(x)=0=>4(5-x)(5-3x)=0$$

Find the critical number(s)

$$x_1=\dfrac{5}{3}, x_2=5$$

$$V(0)=0$$

$$V(5)=0$$

$$V(\dfrac{5}{3})=(10-2(\dfrac{5}{3}))^2(\dfrac{5}{3})=\dfrac{2000}{27}\approx74.1$$

The function $V(x)$ fas the absolute maximum on $[0, 5]$ with value of $\dfrac{2000}{27}$ at $x=\dfrac{5}{3}.$

$$\dfrac{5}{3}\approx1.7$$

$$10-2\cdot\dfrac{5}{3}=\dfrac{20}{3}\approx6.7(in)$$

D. 6.7 in. x 6.7 in. x 1.7 in.; 74.1 in³