Question

The base of a rectangular box is to be twice as long as it is wide. The volume of the box is 864 cubic inches. The material
for the top costs $0.50 per square inch and the material for the sides and bottom costs $0.25 per square inch. Find the
dimensions that will make the cost a minimum.

Find the

length=
width=
height=

Answer 1

Answer:

length = 12 inches

width = 6 inches

height = 12 inches

Step-by-step explanation:

If the length, width, and height are L, W, and H, then:

L = 2W

864 = LWH

If the cost is C, then:

C = 0.50 LW + 0.25 (LW + 2LH + 2WH)

C = 0.50 LW + 0.25 LW + 0.50 LH + 0.50 WH

C = 0.75 LW + 0.50 LH + 0.50 WH

We need to get the cost in terms of a single variable.

Substitute the first equation into the second equation to get H in terms of W:

864 = (2W) WH

864 = 2W² H

432 = W² H

H = 432 / W²

Now substitute the equations for L and H into the cost equation:

C = 0.75 (2W) W + 0.50 (2W) (432 / W²) + 0.50 W (432 / W²)

C = 1.5 W² + 432 / W + 216 / W

C = 1.5 W² + 648 / W

Take derivative and set to 0:

dC/dW = 3W − 648 / W²

0 = 3W − 648 / W²

3W = 648 / W²

3W³ = 648

W³ = 216

W = 6

Therefore:

L = 12

H = 12

Answer 2

The length, width, and height are 12, 6, and 12.

Area of rectangle

It refers to the area of the quadrilateral.

Given

Length(L) = 2 of Width(W)

Height(H)

How to calculate the cost minimum?

$\begin{aligned} LWH &= 864\\ W^{2} H &= 432\\H &= \dfrac{432}{W^{2} } \end{aligned}$

If the cost is c. then

$C = 0.5LW\ +\ 0.25\(LW+ 2WH +2HL)\\C = 0.75LW+ 0.5WH+0.5HL \\C = 0.75*2W^{2} + 0.5W*\dfrac{432}{W^{2} } } + 0.5*\dfrac{432}{W^{2} } *2W\\C = 1.5W^{2} + \dfrac{648}{W}$

So the derivative of C with respect to W

$\begin{aligned} \dfrac{\mathrm{d}C }{\mathrm{d} W} &= 3W - \dfrac{648}{W^{2}}\\W &= 6\\\end{aligned}$

Then

$W=6\\\\L = 2*W=2*6 = 12\\\\H = \dfrac{432}{W^{2} }= \dfrac{432}{6^{2} } =12$

Thus, the length, width, and height are 12, 6, and 12.

More about the Rectangle link is given below.

https://brainly.com/question/10046743