Question

A rectangular box with a square base and no top is being constructed to hold a volume of 120 cm3. The material for the base of the container costs $18/cm2 and the material for the sides of the container costs $3/cm2. Find the dimensions of the cheapest possible box

Answer 1

Base=3.42cm

Height = 10.26 cm

For a total cost of $631.59

Explanation:

In order to solve this problem we can start by drawing what the box will look like. (See attached picture).

So, the problem mentions that the box will have a volume of
`120cm^(3)`. Since the box has a squared base, we can calculate its volume by using the following formula:

`V=b^(2)h`

we can turn this into an equation by substituting the corresponding volume:

`b^(2)h=120`

now, the problem tells us that the base of the box will cost
`$18/cm^(2)` so first, we need to calculate the area of the base of the box. Its area can be calculated by using the following formula:

`A_(base)=b^(2)`

so the cost of the base is calculated with the following equation:

`C_(base)=18b^(2)`

we can find the cost of the sides of the box by following a similar procedure:

`A_(side)=bh`

since there are 4 sides to the box, then we can calculate the total area of the sides by multiplying the formula by 4.

`A_(sides)=4bh`

the problem tells us that the cost of the sides of the container is:
`$3/cm^(2)` so the cost of the sides will be:

`C_(sides)=3(4)bh`

`C_(sides)=12bh`

So the total cost of the box is found by adding the two costs we just found:

`C_(total)=C_(base)+C_(sides)`

`C_(total)=18b^(2)+12bh`

so we can take the volume equation to find an equation we can substitute for h on the cost equation:

`h=(120)/(b^(2))`

when substituting we get:

`C_(total)=18b^(2)+12b((120)/(b^(2)))`

Which simplifies to:

`C_(total)=18b^(2)+(1440)/(b)`

or:

`C_(total)=18b^(2)+1440b^(-1)`

In order to minimize the costs we will now take the derivaative of this function and set it to be equal to zero:

`C'=36b-1440b^(-2)`

`36b-1440b^(-2)=0`

and now we solve for b:

`36b=(1440)/(b^(2))`

`36b^(3)=1440`

`b^(3)=(1440)/(36)`

`b^(3)=40`

`b=\sqrt[3]{40}`

b=3.42 cm

so now we can use this value to find the height:

`h=(120)/(3.42)`

h=10.26 cm

the total cost is found by using the cost equation:

`C_(total)=18b^(2)+1440b^(-1)`

`C_(total)=18(3.42)^(2)+1440(3.42)^(-1)`

`C_(total)=$631.59`