Question

From a thin piece of cardboard 50 in. by 50 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.

Answer 1

Answer:

When dimension of box is 33.33 inches × 33.33 inches ×8.33  then its volume is maximum and is 9259.26 cubic inches.

Step-by-step explanation:

Let h be the length (in inches) of the square corners that has been cut out from the cardboard and that would be the height of the cardboard box.

Since the squares have been cut from cardboard, both sides of the cardboard would reduce by 2h.

Thus, The dimension of box is  (50 – 2h) × (50 – 2h) × h in dimensions.

The volume V of rectangular box = (Length × Breadth × Height) cubic inches.

$V=(50-2h) \times (50-2h) \times h$

$V=(50-2h)^2 \times h$  ..............(1)

Using $(a-b)^2=a^2+b^2-2ab$

$V=h(2500+4h^2-200h)$

$V=2500h+4h^3-200h^2$

For obtaining a box of maximum volume, maximize V as a function of h.

Differentiate both sides with respect to h,

$\frac{dV}{dh}=2500+12h^2-400h$

$\frac{dV}{dh}=4(625+3h^2-100h)$

Solving quadratic equation,$625+3h^2-100h$

$\frac{dV}{dh}=4(3h^2-25h-75h+625)$

$\frac{dV}{dh}=4(h(3h-25)-25(3h-25))$

$\frac{dV}{dh}=4((h-25)(3h-25))$

For maximum, $\frac{dV}{dh}=0$  

thus,$4((h-25)(3h-25))=0$

⇒ $h= 25$ or $h=\frac{25}{3}$

Now check (1) for $h= 25$ and $h=\frac{25}{3}$.

$h= 25$ is not possible as when h is 25 inches then length and breadth becomes 0.

When $h=\frac{25}{3}$.

(1) ⇒ $V=(50-2(\frac{25}{3}))^2 \times\frac{25}{3}=9259.2592593$  

This is the maximum volume the box can assume.

Thus, when dimension of box is 33.3 inches × 33.3 inches ×8.3  then its volume is maximum and is 9259.26 cubic inches.