A company needs to produce a cylindrical can to hold water. It must have a volume of 500ml. The side of the can will cost 40$/m^2 to produce and the top and bottom of the can will cost 50$/m^2 to produce. Find the most economical dimensions.
Find the compound interest and the compound amount of the following #2ooo at 4/2 % compounded annually for 7yr
since pi r^2 h = 500, h = 500/(pi r^2)
the cost function is
c = 40 * 2pi r h + 2*50*pi r^2
= 80 pi r (500/(pi r^2)) + 100 pi r^2
= 40000/r + 100 pi r^2
dc/dr = 200 pi r - 40000/r^2
= 200 (pi r^3 - 200)/r^2
dc/dr = 0 when pi r^3 = 200
To find the most economical dimensions for the cylindrical can, we need to minimize the cost of production while ensuring it has a volume of 500ml.
Let's assume the radius of the base of the can is 'r' and the height of the can is 'h'. The volume V of a cylinder is given by the formula: V = πr^2h.
We need to minimize the cost C of production, which is based on the surface area of the can. The surface area S of a cylinder is calculated using the formula: S = 2πrh + 2πr^2.
In this case, the cost of production includes the cost for the sides, which is $40/m^2, and the cost for the top and bottom, which is $50/m^2.
The cost C of production is given by the formula: C = 40S + 50(2πr^2).
To simplify the calculations, we can rewrite the surface area S in terms of the volume V using the formula for the volume of a cylinder.
Substituting S = (2V/πh) + 2πr^2 into the cost C formula, we get: C = 40[(2V/πh) + 2πr^2] + 50(2πr^2).
Now, we can substitute the given volume of 500ml into the formula for V and simplify further.
V = 500ml = 0.5L = 0.0005m^3
Substituting this value into the cost formula, we get:
C = 40[(2(0.0005)/πh) + 2πr^2] + 50(2πr^2)
= 0.04(0.001/πh + 4πr^2) + 0.1πr^2
= ((0.04/πh) + 0.16πr^2) + 0.1πr^2
To find the most economical dimensions, we need to find the values of 'r' and 'h' that minimize the cost C of production.
To do this, we can minimize the cost function by taking the partial derivatives with respect to 'r' and 'h', equating them to zero, and solving for 'r' and 'h'.
Taking the partial derivative of C with respect to 'r' and 'h' and setting them equal to zero, we get:
∂C/∂r = 0.16π - 0.2πr/h = 0 (Equation 1)
∂C/∂h = (-0.04/πh^2) + 0.16πr^2 = 0 (Equation 2)
Solving Equation 1 for r, we get:
0.16π - 0.2πr/h = 0
0.16π = 0.2πr/h
0.16 = 0.2r/h
Simplifying further, we get:
r/h = 0.16/0.2
r/h = 0.8
Multiplying both sides by h, we get:
r = 0.8h
Substituting this value of r into Equation 2, we get:
(-0.04/πh^2) + 0.16π(0.8h)^2 = 0
(-0.04/πh^2) + 0.512πh^2 = 0
Multiplying through by πh^2, we get:
-0.04 + 0.512πh^4 = 0
Rearranging the equation, we get:
0.512πh^4 = 0.04
Dividing both sides by 0.512 and π, we get:
h^4 = 0.04 / (0.512π)
h^4 = 0.0244/π
Taking the fourth root of both sides, we get:
h = (0.0244/π)^(1/4)
Substituting this value of h back into the expression r = 0.8h, we get:
r = 0.8 * (0.0244/π)^(1/4)
Therefore, the most economical dimensions that minimize the cost of production while ensuring a volume of 500ml are:
r = 0.8 * (0.0244/π)^(1/4)
h = (0.0244/π)^(1/4)
By substituting these values into the volume formula, we can verify that the volume of the cylindrical can will indeed be 500ml.