What are the dimensions that minimizes the cost of the box.?

Question

Rectangular boxes with a volume of 8 ft³ are to be made of three materials. The material for the bottom of the box costs $12 per ft², the top of the box costs $4 per ft² and the material for the sides costs $10 per ft². What are the dimensions of the box that minimizes the cost of the box? Show all work. Round final dimensions of the box to the nearest tenth of a foot. (1 decimal place)

Answer 1

When l eliminate hieght , find cost as function of length and width, use partial derivatives, set to zero and simulataneously solve, I find the top and bottom must be square.

given the top and bottom are square,

$=16w²+320/w where h=8/w²

d$/dw=32w-320/w²

w³=10 , w=10^1/3 when d$/dw=0

d²$/dw²=32+620/w³ is positive for w³=10 so the point is a minimum

h=8/10^2/3

Answer 2

Let x be the length, y be the width and z be the height of the rectangular box.

Since the volume is 8 ft³ then xyz=8 and then z= 8/(xy)

Therefore

1. Length = x
2. Width = y
3. Height = 8/(xy)

Then the Cost as a function x,y is given by

C(x,y) =12 xy +2[ 8/(xy)] (x +y)⋅10 +4xy.

Now minimize C(x,y) using the Second Derivative test.

I hope you will be done in a couple of hours till I will be back

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C(x,y) =16 xy +[160/(xy)] (x +y) = 16xy +160/ x + 160/y

C(x,y) = 16xy +160/ x + 160/y

Then suffice to minimize

Ω= xy + 10/ x +10/y

Ω_X= y -10/ x²

Ωy= x -10/ y²

Let us solve the system

y -10/ x² = 0

x -10/ y² = 0

to find the critical points.

The solution being (x,y) = (10 ^1/3, 10^1/3) the only critical point.

SECOND DERIVATIVE TEST

Ω_xx = 20/ x³

Ω_yy = 20/y³ Ω_xy= Ω_yx =1

Then Ω_xxΩ_yy - Ω_xy² = 3 , at (10 ^1/3, 10^1/3)

and Ω_{xx ,} Ω_yy are both positive at (x,y) = (10 ^1/3, 10^1/3) , so the function Ω(x,y) and naturally

C(x,y) have a minimum at (x,y) = (10 ^1/3, 10^1/3)

Then the dimensions of the box are as

x = 10 ^1/3

y= 10^1/3

z= 8/(10^2/3)