An aquarium is to be constructed to hold 2160 in3. The base is to made of slate and the sides of glass. If slate costs 5 times as much as the glass per sq in, find the dimensions that will minimize the cost of constructing the aquarium.

I started going about this problem by figuring out what is given. Height = 15 , base = 12 x 12....C = 5xy + 2 xz + 2 zy ; xyz = 2160...base is x by y

C(x,y) = 5 x y + 2 (2160) / y + 2 (2160)/x

Is this correct thus far?

How do we know h = 15 ?

If we do not know h, then the solution is something that looks like a space needle :)

L * w * h = 2160
if h = 15
L * w = 144
L = 144/w

cost = c = 5(L*w) + 2 L*h + 2 w*h
c = 5 L w + 2 w h + 2 L h
if we know h = 15, then
c = 5 L w + 30 w + 30 L
c = 5*144 + 30 w + 30 (144/w)
dc/dw = 0 for min
= = 0 + 30 - 30(144)/w^2
w = 12

Yes, you have made a good start. You correctly identified the given information and the objective function for the cost of constructing the aquarium. The objective function C(x, y) represents the total cost in terms of the dimensions x and y of the base of the aquarium.

To continue, you need to find the dimensions that will minimize the cost of constructing the aquarium. This can be done by taking the partial derivatives of C(x, y) with respect to x and y, setting them to zero, and solving the resulting equations simultaneously to find the critical points.

Let's find the partial derivative of C(x, y) with respect to x:

∂C/∂x = 0 + 0 + 2(2160)/x^2
= 4320/x^2

Next, let's find the partial derivative of C(x, y) with respect to y:

∂C/∂y = 5 + 2(2160)/y^2
= 5 + 4320/y^2

Setting both partial derivatives to zero, we have:

∂C/∂x = 4320/x^2 = 0
=> x^2 = 4320
=> x = √4320

∂C/∂y = 5 + 4320/y^2 = 0
=> 4320/y^2 = -5
=> y^2 = -4320/5
=> y^2 = -864
This equation has no real solutions, which is not physically meaningful in this context. Therefore, there are no critical points with respect to y that need to be considered.

Now that we have the critical point x = √4320, we can substitute it back into the objective function C(x, y) to find the minimum cost:

C(√4320, y) = 5y + 2(2160)/y + 2(2160)/(√4320)
= 5y + 4320/y + 864/√4320

Hence, the dimensions that will minimize the cost of constructing the aquarium are x = √4320 (approximately 65.73) and y can be any value (since there are no critical points with respect to y).