Cost optimization problem?

**softballchick** · January 27th 2011, 07:11 PM

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?

**pickslides** · January 27th 2011, 07:21 PM

Did it on the fly, but I get $\displaystyle C = 26w^2+\frac{90}{w}$

**Prove It** · January 27th 2011, 07:32 PM

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...

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

$\displaystyle 30 = 2w^2 h$

$\displaystyle 15 = w^2h$

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

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

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

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

Now minimise the cost function.

**softballchick** · January 28th 2011, 07:02 AM

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

Now what? where do I plug this into?

**Unknown008** · January 28th 2011, 08:11 AM

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

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

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