A holding pen for fish is to be made in the form of a rectangular solid with a square base and open top. The base will be slate that costs $4 per square foot and the sides will be glass that costs $5 per square foot. If the volume of the tank must be 50 cubic feet, what dimensions will minimize the cost of construction?
The answer is 5 X 5 X 2, but I don't understand how to get that.
Assuming that length = x, width = x. height = h, these are the two equations I came up with:
(x^2)h = 50
SA = x^2 + 4xh
Am I doing this right?
You are doing fine so far.
For these kind of max/min questions, look for "something" which will be either maximized or minimized. What ever that "something" is , you will need an equation that says
"something" = ......
In this case I see, " .... what dimensions will minimize the cost of construction? "
So in this case it is the COST
cost = 4(x^2) + 5(4xh)
= 4x^2 + 20xh
but we know:
x^2 h = 50
so h = 50/x^2
cost = 4x^2 + 20x(50/x^2)
= 4x^2 + 1000/x
now differentiate with respect to x
d(cost)/dx= 8x - 1000/x^2
= 0 for a min of cost
8x = 1000/x^2
8x^3 = 1000
take cube root of both sides
2x = 10
x = 5
then h = 50/5^2 = 2
and there you have it!
base is 5 by 5 , and the height is 2
Thanks so much Reiny!
I never thought of adding the cost to the equation, but now I know.
Yes, you are on the right track! Let's go step by step to find the solution.
You correctly set up the two equations. The first equation represents the volume of the tank: (x^2)h = 50. The second equation represents the surface area of the tank, considering the top is open: SA = x^2 + 4xh.
To minimize the cost of construction, we need to minimize the total surface area (SA) while keeping the volume (V) fixed at 50 cubic feet.
We can start by expressing h in terms of x using the volume equation: h = 50/(x^2).
Substituting this expression for h in the surface area equation gives us: SA = x^2 + 4x(50/(x^2)).
Now, we need to simplify this equation further. Multiplying 4x by 50 gives us 200x/x^2, which simplifies to 200/x.
The surface area equation becomes: SA = x^2 + 200/x.
To find the minimum of SA, we can take its derivative with respect to x, set it equal to zero, and solve for x. So, let's find dSA/dx:
dSA/dx = 2x - 200/x^2.
Setting this derivative equal to zero, we get: 2x - 200/x^2 = 0.
To make solving this equation easier, we can multiply both sides by x^2: 2x^3 - 200 = 0.
Now, solving for x^3: 2x^3 = 200 ==> x^3 = 100 ==> x = cube root of 100.
Approximating the cube root of 100, we get x ≈ 4.6416.
Since the dimensions must be positive, we can round up x to the nearest whole number, which gives us x = 5.
Now that we have x = 5, we can substitute it back into the volume equation: (5^2)h = 50.
Solving for h, we get h = 50/25 = 2.
Therefore, the dimensions that minimize the cost of construction are 5 x 5 x 2.
It's important to note that in this case, we are assuming the base and sides have the same material. If they had different materials, the cost optimization might involve a different approach.