To construct a tincan, V=32pi m^3, The cost per square meter of the side is half of the top and bottom of can. What are the dimensions and the cost?
V=πr²h=32pi SA=2πr²+2πrh
h=32/r²
Domain={r>o}
Let x be the cost, I subbed 32/r² for h
C=f(r)=2πxr²+(32xπ/r)
f'(r)=4πxr + 32xπ/r²
common denominator
f'(r)=(4πxr³+ 32xπ)/r²
I factored, it, r=-2, which is not in the domain.
Where did it go wrong?
Cost = 2(2πr^2) + 1(2πrh)
= 4πr^2 + 2πr(32/r^2)
= 4πr^2 + 64π/r
d(Cost)/dr = 8πr - 64π/r^2 = 0 for a min of Cost
8πr = 64π/r^2
r^3 = 8
r = 2
h = 32/4 = 8
Can should have a radius of 2 and a height of 8
I did the derivative wrong! Thanks!
It seems like you encountered an error while calculating the derivative of the cost function. Let's go through the steps again to identify where the mistake occurred.
First, let's declare the variables and equations given in the problem:
- Volume: V = 32π m^3
- Surface Area: SA = 2πr² + 2πrh (where r is the radius and h is the height)
- Cost per square meter of the side: x
- Cost per square meter of the top and bottom: 2x
To begin, we need to express h in terms of r in order to eliminate one variable from the equation. Since V = πr²h, we can rearrange this equation to solve for h:
h = V / (πr²)
h = (32π) / (πr²)
h = 32 / r²
Now, we can write the cost function in terms of r and x:
C(r) = 2πxr² + 2πrh
Substituting the value of h:
C(r) = 2πxr² + 2πr(32/r²)
C(r) = 2πxr² + 64π/r
Next, if we want to find the optimum value of r that minimizes the cost, we need to find the derivative of the cost function with respect to r:
C'(r) = 4πxr + (64π/-r²)
Now, let's simplify the derivative expression:
C'(r) = 4πxr - 64π/r²
To find where the derivative equals zero (critical points), we set C'(r) equal to zero and solve for r:
0 = 4πxr - 64π/r²
4πxr = 64π/r²
4πr³ = 64π
r³ = 16
r = cuberoot(16)
r = 2
Therefore, the radius of the tin can is 2 meters.
Now, let's calculate the cost by substituting r = 2 back into the cost function:
C(r) = 2πx(2)² + 64π/(2)
C(r) = 8πx + 32π
C(r) = 8πx + 32π
Since the cost per square meter of the side is given as half of the top and bottom, we can set up the equation:
x = (1/2)(2x)
x = x
This shows that x equals itself and does not provide any helpful information about the cost.
In summary, the dimensions of the tin can are a radius of 2 meters, and the cost cannot be determined based on the information provided. It appears there may be an error in the problem statement regarding the cost per square meter.