Maximizing Volume and Total Cost Help Please

[Getupjb]

A building with side x and height y
The volume of the building is V = x^2 y
The total cost of the building will be T = ax^2 + by
T, a, and b are constants
I need to derive a formula for x and y that will maximize the volume for a a fixed amount of the cost.
x will have 2 values, one of which is zero.

According to my practice problem I should take the Total cost and find a value for y.
Then plug the y value into the volume equation.
Is that process correct?
So y = T/ax^2+b?
...Im so lost, i would really appreciate any help you could give me.
Thanks

 

[galactus]

Try solving the T equation for x^2.

\(\displaystyle x^{2}=\frac{T-by}{a}\)

Sub into the volume equation:

\(\displaystyle V=\frac{T-by}{a}y\)

Differentiate with respect to y:

\(\displaystyle \frac{dV}{dy}=\frac{T-2by}{a}\)

Set to 0 and solve for y:

After you find y, use it to find x and the volume with respect to T,a, and b.

 

[Getupjb]

Thanks for your help galactus, I really appreciate it.

So then

0=T-2by/a 2by=T/a y=T/2ba

and

V = x^2 * T/2ba

derive dV/dy = 2x * T/2ba set equal to 0

0 = 2x * T/2ba -2x = T/2ba x = T/-ba

Honestly Im not sure what the answer will look like at all.
So if im way off, please let me know.

Can i plug my x & y into the Volume equation?

V = (T/-ba)^2 * T/2ba ?

Thanks again for the help

 

[Gene]

You went astray with
0=T-2by/a 2by=T/a y=T/2ba
Which you should have written as
0=(T-2by)/a
When you multiply by a you get
0=T-2by
Solving for y
T=2by
y=T/2b
Galactus already gave you dV/dy, the change in volume as y changes. You don't need to do it again.
Plug y into the Total cost equation and solve for x.