Max-Min Problems: Optimizing Fencing, Fare Charges, and Can Construction

[courtrigrad]

Hello all

I have a few questions on maxima and minima

1. A farmer plans to fence in a rectangular pasture located adjacent to a river. She has 640 yards of fencing available and will need to use double fencing on the two opposite sides. What dimensions should be used so that the enclosed area will be maximum? (N0 fencing on the river)

Would the function be x(320 - 2x)?



2. If 40 passengers hire a special car on a train, they will be charged $8.00 each. For each passenger over the 40 this fare is cut $0.10 apiece for all passengers. How many passengers will produce th greatest income for the railroad?

I am not sure about the equation for this one.

3. A manufacturer wishes to construct a cylindrical can to hold 100 pi in^3. The top and bottom of the can are to be stronger than the sides. The tin used in making the top and bottom will cost 2.5 cents per square inch while the metal used in making the sides will cost 1.35 cents per square inch. What dimensions should be used to minimize the cost of the metal?

All I know it that the area of a cylinder = pi*r^2 * h.

Any help is greatly appreciated

Thanks

 

[Gamma]

1) Area = 2x(320 - 2x). So you are right.

2) confusing at first. but as i understand, assuming the number of people who will bring maximum income to be N (N>40),
I = [8 - 0.1(N - 40)] * N

3) Area of the sides = 2 pi r * h

Areas of the top and bottom = 2 * pi r^2
So find the total cost in terms of r. You have to subtitute for h from the given condition that the volume pi r^2 h = 100 pi

regards,
gamma

 

[courtrigrad]

thanks a lot!

 

[courtrigrad]

for #3 would the equation be:

5 pi r (100 / r^2) + 2.7pi r h?

thanks

 

[Gamma]

I don't know how you came up with this. I get
5 pi r^2 + 270 Pi/r