A cylindrical oil-storage tank is to be constructed for which the following costs apply:
cost per square meter metal for ides $30.00
combined costs of concrete base and metal bottom $37.50(cost per square meter)
top 7.50 (cost per square meter)
The tank is to be constructed with dimensions such that the cost is minimum for whatever capacity is selected.
a) One possible approach to selecting the capacity is to build the tank large enough for an additional cubic meter of capacity to cost $8.(note that this does not mean $8 per cubic meter average for the entire tank.) what is the optimal diameter and optimal height of the tank?
The last part of the question confuses me, I will solve the question in the traditional way
Let the radius be r m, the height be h m and the volume be V m^3, where V is a constant
We know V = πr^2h
h = V/(πr^2)
Cost = 7.5(top area) + 37.5(bottom area) + 30(side area)
= 7.5πr^2 + 37.5πr^2 + 30(2πr)(h)
= 45πr^2 + 60V/r
d(Cost)/dr = 90πr - 60V/r^2 = 0 for a min of Cost
90πr = 60V/r^2
r^3 = 2(πr^2h)/(3π)
3r = 2h
The ration of radius to height has to be 2 : 3
I will let you deal with the weird condition at the end.
After doing a "search" for your question, I noticed that
MathMate had answered the same question in a much more detailed way.
http://www.jiskha.com/display.cgi?id=1289839685
To find the optimal diameter and optimal height of the tank, we need to minimize the total cost of the tank, taking into account the cost of the metal for the sides, the combined cost of the concrete base and metal bottom, and the cost of the top.
Let's express the dimensions of the tank in terms of radius (r) and height (h). The formula for the cost of the metal for the sides is given as $30.00 per square meter, so the cost for the sides can be calculated as:
Cost of metal for sides = 2πrh * $30.00
The cost of the base and bottom is given as $37.50 per square meter, and since the base is circular, the cost for the base can be calculated as:
Cost of base = πr^2 * $37.50
The cost of the top is given as $7.50 per square meter, and since the top is also circular, the cost for the top can be calculated as:
Cost of top = πr^2 * $7.50
Now, we can express the total cost (C) of the tank as a function of r and h:
C = 2πrh * $30.00 + πr^2 * $37.50 + πr^2 * $7.50
Since we're tasked with selecting the capacity such that the cost for an additional cubic meter of capacity is $8, we can express the capacity (V) of the tank as:
V = πr^2h
Let's rewrite the total cost function in terms of V:
C = 2V * $30.00 + V/r * $37.50 + V/r * $7.50
C = 2V * $30.00 + V * $45.00/r
To optimize the cost, we need to find the values of r and h that minimize C. Since we have an equation for C in terms of V, we can find the values of r and h that minimize C by finding the values of r and h that minimize V.
Now, we have the additional information that an additional cubic meter of capacity costs $8. This means that the marginal cost, dC/dV, is equal to $8:
dC/dV = $8
Differentiating the cost function with respect to V:
dC/dV = 2 * $30.00 + $45.00/r = $8
Simplifying the equation:
2 * $30.00 + $45.00/r = $8
60 + 45.00/r = 8
45.00/r = -52
r = -45.00/52
The radius, r, cannot be negative, so there is no solution for this equation. Hence, there is no minimum cost solution for this particular scenario.
It's important to note that the optimal diameter and optimal height of the tank may vary depending on the specific constraints and cost conditions given.