Thread: Cost optimization problem?

1. Cost optimization problem?

A rectangular storage container with an open top is to have a volume of 30 cubic meters. The length of its base is twice the width. Material for the base costs 13 dollars per square meter. Material for the sides costs 9 dollars per square meter. Find the cost of materials for the cheapest such container.
Total cost = ________________dollars

i got about $234. i am pretty sure I did all the steps right. First I got the equation which is C=13Lw+6LW+6WH got it to one variable 26w^2 + 18w(30/2w^2) derived and set equal to zero Found W and then plugged it in. Did I do something wrong? 2. Did it on the fly, but I get$\displaystyle \displaystyle C = 26w^2+\frac{90}{w}$3. Originally Posted by softballchick A rectangular storage container with an open top is to have a volume of 30 cubic meters. The length of its base is twice the width. Material for the base costs 13 dollars per square meter. Material for the sides costs 9 dollars per square meter. Find the cost of materials for the cheapest such container. Total cost = ________________dollars i got about$234.

i am pretty sure I did all the steps right.

First I got the equation which is C=13Lw+6LW+6WH

got it to one variable

26w^2 + 18w(30/2w^2)

derived and set equal to zero

Found W

and then plugged it in.

Did I do something wrong?
Let's see...

$\displaystyle \displaystyle V = 30 = lwh$ and $\displaystyle \displaystyle l = 2w$, so

$\displaystyle \displaystyle 30 = 2w^2 h$

$\displaystyle \displaystyle 15 = w^2h$

$\displaystyle \displaystyle h= \frac{15}{w^2}$.

The area of the base is $\displaystyle \displaystyle lw = 2w^2$ and the area of the sides are $\displaystyle \displaystyle 2lh + 2wh = \frac{60}{w} + \frac{30}{w} = 90w^{-1}$.

Since the cost is $\displaystyle \displaystyle \$ 13$per square metre for the base and$\displaystyle \displaystyle \$9$ per square metre for the sides,

$\displaystyle \displaystyle C = 13\cdot 2w^2 + 9\cdot 90w^{-1} = 26w^2 + 810w^{-1}$.

Now minimise the cost function.

4. I minimized it and got w=(810/52)^(1/3)

Now what? where do I plug this into?

5. Now you plug this into your original equation for the cost, that is:

$\displaystyle \displaystyle C = 26w^2 + \dfrac{810}{w}$