Math word problem-calculus please?
A cylindrical can is to hold 4π cubic units of juice. The cost per square unit of constructing the metal top and the bottom is twice the cost of constructing the cardboard side. what are the dimensions of least expensive can?
Please use calculus logically to solve this problem and please show me your work. Thank you!
Comments
The total surface area of the can is:
A = 2πrh + 2πr²
The volume of the can is:
V = πr²h = 4π, then h = 4/r²
so that; A = 2πr(4/r²) + 2πr²
We want the area of the top to be a minimum while
having the same volume. Therefore, we need to minimize
r and vary the height. We will take the derivative of the
total surface to accomplish this aim.
dA/dr = 8π(-1/r²) + 4πr = 0,
r = ³√2,
and since the cost of Area of top is twice the area of cardboard; that is,
2πr² = 2(2πrh), then h = r/2
The dimensions for minimum material cost is then:
r = ³√2, and h = 2 ³√2
a million. dZ/dt = ?Z/?x·dx/dt = [R/(R+x)]²·dx/dt dZ/dt(x=a million.05) = [3/(3+a million.05)]²·a million.40 5 = 0.eighty $/min Is the expression for Z (Rx/R)+x or Rx/(R+x)? 2. ?V = ?V/?r·?r + ?V/?l·?l + ?V/?p·?p + ?V/?n·?n = ?/8[(4pr³/nl)(a million.05r) + (-pr^4/nl²)(a million.05l) + (r^4/nl)(0.9p) + (-pr^4/n²l)(0.7n)] = ?/8[4.2(pr^4/nl) - a million.05(pr^4/nl) + 0.9(pr^4/nl) - 0.7(pr^4/nl)] = ?/8[3.35(pr^4/nl)] = 3.35(?pr^4/nl) = 3.35V 335% strengthen