1. ## Optimization Problem

A rectangular storage container with an open top is to have a volume of 10 m3. The length of this base is twice the width. Material for the base costs $10 per square meter. Material for the sides costs$6 per square meter. Find the cost of materials for the cheapest such container.

I understand the calculus for the most part, but creating the initial formula for this is confusing me to no end. So far I've got: Cost = 10(b^2) + 4(6b*h) where b=l*w where l=2w so b = 2w*w = 2w^2.

Now I assume I mess up somewhere when I go with 10m cubed = (l*w)*h. So 10 = b*h. b = 2w^2, So h = 10/2w^2. I proceed to throw that into the Cost equation so that C = 10(2w^2)^2 + 24(2w^2)(10/2w^2). Which simplifies to 10(2w^2)^2 + 240.

C' = 20(2w^2) * 4w = 10w^3

Not quite sure where I messed up, as the problem is kinda long. I would appreciate it if someone could point out my error so I could work on it. I have trouble with these optimization problems as they are mostly word problems that tend to confuse me.

2. The first term in your cost formula is wrong: if b = lw, then b has units m^2, so b^2 has units m^4, so 10 dollars/m^2 * b^2 has units dollars * m^2...

If l and w are the length and width of the base, and h is the height of the sides, you must have $lwh = 10$, $l=2w$, and your cost function is $C := 10 lw + 6(2l+2w)h$ (if I have apprehended you correctly). Can you go from there?

3. Originally Posted by abel2
A rectangular storage container with an open top is to have a volume of 10 m3. The length of this base is twice the width. Material for the base costs $10 per square meter. Material for the sides costs$6 per square meter. Find the cost of materials for the cheapest such container.

I understand the calculus for the most part, but creating the initial formula for this is confusing me to no end. So far I've got: Cost = 10(b^2) + 4(6b*h) where b=l*w where l=2w so b = 2w*w = 2w^2.

Now I assume I mess up somewhere when I go with 10m cubed = (l*w)*h. So 10 = b*h. b = 2w^2, So h = 10/2w^2. I proceed to throw that into the Cost equation so that C = 10(2w^2)^2 + 24(2w^2)(10/2w^2). Which simplifies to 10(2w^2)^2 + 240.

C' = 20(2w^2) * 4w = 10w^3

Not quite sure where I messed up, as the problem is kinda long. I would appreciate it if someone could point out my error so I could work on it. I have trouble with these optimization problems as they are mostly word problems that tend to confuse me.

See the attached picture for what I did.