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?:(

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.

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

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.