Question

The volume of an open-top rectangular box is 4500 cc (cubic centimeters). The length of the rectangular base of the box is twice the width. What height will make the surface area as small as possible?

Answer 1

Answer:

The height that make the surface area as small as possible is 10 cm

Step-by-step explanation:

Let

L -----> the length of the base of the box

W -----> the width of the base of the box

H ----> the height of the box

we know that

The volume of the box is equal to

$V=LWH$

we have

$V=4,500\ cm^3$

so

$4,500=LWH$ ----> equation A

$L=2W$ ------> equation B

substitute equation B in equation A

$4,500=(2W)WH$

$4,500=2W^2H$

Solve for H

$H=\frac{2,250}{W^{2}}$ -----> equation C

The surface area of a open box is equal to

$SA=2(L+W)H+LW$ ----> equation D  

Substitute equation B and equation C in equation D

$SA=2(2W+W)(\frac{2,250}{W^{2}})+(2W)W$

$SA=6W(\frac{2,250}{W^{2}})+2W^2$

$SA=6(\frac{2,250}{W})+2W^2$

$SA=\frac{13,500}{W}+2W^2$

Convert to function notation

$f(W)=\frac{13,500}{W}+2W^2$

Find the derivative

$f'(W)=-\frac{13,500}{W^2}+4W$

set equal to zero

$f'(W)=0$

$0=-\frac{13,500}{W^2}+4W$

Solve for W

$\frac{13,500}{W^2}=4W$

$W^3=\frac{13,500}{4}$

$W=15\ cm$

Find the value of L (equation B)

$L=2(15)=30\ cm$

Find the value of H (equation C)

$H=\frac{2,250}{15^{2}}=10\ cm$

therefore

The height that make the surface area as small as possible is 10 cm