Optimization of the surface area of a open rectangular box to find the cost of materials

Question

A rectangular storage container with an open top is to have a volume of 10 cubic meters. The length of the box is twice its width. Material for the base costs ten dollars per square meter and for the four sides the cost is six dollars per square meter. Find the cost of materials for the cheapest such container.

I am having issues finding the constraint and expression to optimize. How do I go about finding them? I know how to solve the problem once I find those equations but I am unable to. Thanks!

Answer 1

So we know $L = 2w$ and $V = Lwh$. So we have $V = 2w^{2}h = 10$.

And so the cost of materials is $10b + 6*2*wh + 6*2*Lh$. So that's \$10 for the base and six dollars for the four sides. Two of the sides are $wh$ in area, and the other two are $Lh$ in area.

And so we want to $min$ $10 * 2w^{2} + 12wh + 24w^{2}h$ such that $w^{2}h = 5$. This gives us $w = \sqrt{ \frac{5}{h} }$.

So we are minimizing $20w^{2} + 12wh + 120$ with respect to $w$.

Can you take it from here?

Answer 2

$L$ = length, $W$ = width, $H$ = height, $V$ = volume, $C$ = cost

As established in your constraints, $L = 2W$

$V = LWH = 10$

$C = 10LW + 6(2LH + 2WH) = 10LW + 12LH + 12WH$

The above comes from the base being $LW$ with a cost of ten. The walls are $LH$ and $WH$, of which there are 2 each with a cost of 6.

Using the relationship between length and width, you can start substituting variables.

$V = (2W)WH = 2HW^2 = 10$

$C = 10(2W)W + 12(2W)H + 12WH = 20W^2 + 24WH + 12WH = 20W^2 + 36WH$

Now you have a relation for width and height, so you can use that to sub as well.

$2HW^2 = 10$

$H = 10/(2W^2) = 5/W^2$

$C = 20W^2 + 36WH = 20W^2 + 36W(5/W^2) = 20W^2 + 180/W$

From there you find the minimum of the equation where $W > 0$