Optimisation problem with multiple contraints

[Modine]

An underground cistern is to be constructed to hold 100 cubic metres of radioactive waste. The cistern is to be a circular cylinder in shape. The circular base and vertical sides, which are all underground, cost $100 per square metre and the lid at ground level, costs $300 per square metre because of the necessary shielding. Furthermore, the depth of the tank cannot exceed 6m because of the hard rock layer beneath the surface, which would increase the excavation costs enormously if it were to be penetrated. Finally, the radius of the tank cannot exceed 4m, because of space limitations. What dimensions for the tank will keep its cost to a minimum?

(The answer answer is radius = 2.3m and height of 6m.)

I've set up the equations:
Cost=300πr^2 +100(πr^2+2πrh)
100=πr^2h
So Costs =300πr^2 +100(πr^2+2/r)
dC/dr=(800(πr^3-25))/r^2

Solving that I'm getting an answer of 1.99m and a height of 8.03m, which is greater than 6m. I'm stumped, how do I factor in or consider the constrictions for radius and height?

Thanks in advance

 

[stapel]

An underground cistern is to be constructed to hold 100 cubic metres of radioactive waste. The cistern is to be a circular cylinder in shape. The circular base and vertical sides, which are all underground, cost $100 per square metre and the lid at ground level, costs $300 per square metre because of the necessary shielding. Furthermore, the depth of the tank cannot exceed 6m because of the hard rock layer beneath the surface, which would increase the excavation costs enormously if it were to be penetrated. Finally, the radius of the tank cannot exceed 4m, because of space limitations. What dimensions for the tank will keep its cost to a minimum?

(The answer answer is radius = 2.3m and height of 6m.)

I've set up the equations:
Cost=300πr^2 +100(πr^2+2πrh)

Some steps are missing here. I think you started with the areas of the base, the top, and the sides:

. . .area of top, bottom: A_circ = (pi)r²
. . .area of sides: A_cyl = 2(pi)rh

Then you brought the costs into play, multiplying each of the areas by the relevant cost:

. . .cost of top: C_top = 300((pi)r²)
. . .cost of remainder: C_under = 100(2(pi)rh + (pi)r²)

So the total cost is:

. . .C = 300(pi)r² + 100[2(pi)rh + (pi)r²]

But this gives the cost in terms of two variables, and you need this to be in terms of one variable only, so you looked at the volume information.

100=πr^2h

The volume is given as being fixed at V = 100, and the shape of this volume is a right circular cylinder, so:

. . .Vcyl = (pi)r²h = 100

Solving this for h in terms of r (so you can eliminate a variable), you got:

. . .100/((pi)r²) = h

You plugged this into the cost equation to get:

So Costs =300πr^2 +100(πr^2+2/r)

But I'm not seeing how you took:

. . .2(pi)rh = 2(pi)r * (100/(pi)r²) = [2 * 100 * pi * r] / [pi * r²]

...and got 2/r...?