A containing tank is to be constructed that will hold oil when filled. The shape of the tank is to be a cylinder (including a base) surmounted by a cone (i.e., with a conical top). The cone's height is always equal to its radius which is also the radius of the cylinder. The cost to construct the cylinder portion of the tank is $300 per square meter while the cost for the cone portion is $400 per square meter. Write a function that takes two inputs (TotalVolume and r) and calculates and displays the total cost of constructing the tank. For this problem, when TotalVolume = 500 and r=5.0, the result should be: $112,280. Use the variables listed below. “pi” is a predefined number. Name your function Tank1.m
TotalVolume = Vcone + Vcylinder
Vcone = 1/3(pi*r*r*r)
Vcylinder = pi*r*r*h, where h is height of the cylinder
SAcone = pi*r*r*sqrt(2)
SAcylinder = 2*pi*r*h
SAcylinderbase = pi*r*r
Add the following feature to your solution – instead of entering the radius, solve for the optimum radius and plot the cost as a function of the radius, stepping the radius from 1.0 – 7.0 in steps of 0.1. For this problem, TotalVolume = 500. Note that TotalVolume should still be an input to your function, and that you should use 500 as your input value when you call it in the command window. Output the optimal radius and the lowest cost it produces. Hint: you may want to store the radii values in an array and the output from your function in a second array. Name your function Tank2.m.
Add the following feature to your solution – change the cost structure to allow discounts as follows – for the cylindrical portion of the tank, the cost is $300 per square meter up to and including 200 square meters. Any part of the cylindrical surface area in excess of 200 square meters gets discounted 10%. Similarly, the cost of the conical portion of the tank is $400 per square meter up to and including 100 square meters. Any part of the conical surface area in excess of 100 square meters gets discounted 20%. Note that the discount is only applied to the excess surface area. As in the previous step, solve for and plot the cost as a function of the radius for a volume of 500, stepping the radius from 1.0 – 7.0 in steps of 0.1. Note that TotalVolume should still be an input to your function, and that you should use 500 as your input value when you call it in the command window. Output the optimal radius and the lowest cost it produces. Name your function Tank3.m.
Now, create a new function by placing Tank3.m in a loop with user inputs for minimum volume, maximum volume, and volume increment, i.e., 500, 1000, 100, and output the optimal radius and cost at that radius. Make a single plot showing the curves for all the volumes – each curve representing a single distinct volume. Name your function Tank4.m.
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- Submitted On 17 Dec, 2014 09:34:55