Help regarding optimization problem

Question

In order to construct a box whose base length is 3 times the base width with volume of 50 ft^3, using materials to build the top and bottom that cost $ 10 per ft^2 and $ 6 per ft^2 for the sides. Determine the dimensions that will minimize the cost to build the box.

Answer 1

Let's start by defining some variables:

h: height

w: width

l: length

 

The cost of the top of the box will be:

top_cost = $10/ft² * w*l

 

The cost of the bottom of the box will be the same:

bottom_cost = $10/ft²* w*l

 

The cost of the sides of the box will be:

sides_cost = $6/ft² * h*w + $6/ft² * h*w + $6/ft² * h*l + $6/ft² * h*l

 

sides_cost = 12hw + 12hl

 

The total cost is the sum of those costs:

total_cost = top_cost + bottom_cost + sides_cost

 

total_cost = 20lw + 12hw + 12hl

 

Now let's write two of the variables with respect to the third one. First it is given that the length is 3 times the base, so:

l=3w.

 

Next, it is given that the volume is 50 ft³, i.e. V = lwh

50 = 3w²h

h=50/(3w²)

 

We now have the length and the height defined with respect to the width. We then plug those expressions into the total_cost expression and you should get:

 

total_cost = 60w² + 800/w

 

Now that we have the total_cost expressed with only one variable (the width), we can differentiate it, with respect to that variable.

 

d(total_cost)/dw = 120w - 800/w²

 

To find the minimum, we set that equation to 0 and we solve for w:

 

120w - 800/w² = 0

 

which gives you: w= 1.88 ft

 

Now that you have one dimension, you can figure out the other two dimensions with the expressions found earlier.