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WHITEO · Dec 29, 2006, 06:44 AM

A tank with a rectangular base and rectangular sides is to be open a the top. It is to be constructed so that its width is 4 meters and its volume is 36 cubic meters. If building the tanks costs $10 per square meter for the base and $5 per square meter for the sides, what is the cost of the least exensive tank?:(

galactus · Dec 29, 2006, 07:37 AM

It's nice they give a dimension, otherwise, it'd be a little more involved.

The volume of the box is $xyz=36\Rightarrow{4xz=36}$ ... [1]

The surface area is: $S=2xz+2\cdot{4z}+4x=\underbrace{2xz}_{\text{sides} }+\underbrace{8z}_{\text{ends}}+\underbrace{4x}_{\ text{base}}$

Incorporate the cost of the base and sides:

$C(x)=10xz+40z+40x$ ... [2]

Solve [1] for z and sub into [2]:

$10x(\frac{9}{x})+40(\frac{9}{x})+40x=40x+\frac{360 }{x}+90=4x+\frac{36}{x}+9$

Differentiate:

$\frac{d}{dx}[4x+\frac{36}{x}+9]=4-\frac{36}{x^{2}}$

Set to 0 and solve for x:

$4-\frac{36}{x^{2}}=0$

Can you finish now? Find x. You have y=4. All you need then is z.

hannah94 · Dec 11, 2012, 02:34 PM

Do I just fill in the x value I found into the c(x) equation and then take the derivative and solve for z?

ebaines · Dec 12, 2012, 08:04 AM

Originally Posted by hannah94

Do I just fill in the x value I found into the c(x) equation and then take the derivative and solve for z?

No.

Once you have values for x and y, substitute them into the first equation $xyz=36$ and solve for z.