# I have two questions about Applied Optimization in Calculus; 1) What are the dimentions of the lightest open-top right circular cylendrical can that will hold a volume of 1000 cm^3?   2) a rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a single-strand electric fence. With 800 m of wire at your disposal, what is the largest area you can enclose, and what are the dimentions?

In both cases the idea is to solve for one of the variables in terms of the other:

(a) The volume of a right cylindrical can is `V=pir^2h` ; in this case we are given V=1000

Since the can is open-topped, the area of the can is `A=pir^2+2pirh` . We would like to minimize the area (assuming the can is made of uniform thickness material, the weight of the can is proportional to the area.)

Solve `1000=pir^2h` for h in terms of r: `h=1000/(pir^2)`

Then `A=pir^2+2pir(1000/(pir^2))`

`A=pir^2+2000/r` We want to minimize this, so we take the derivative with respect to r and find the zero of the derivative.

`(dA)/(dr)=2pir-2000/r^2`  Then

`2pir-2000/r^2=0==>pir^3=1000`

`r=10/root(3)(pi)~~6.828`  `h=1000/(pir^2)=1000/(pi(10/root(3)(pi))^2)~~6.828`

So the dimensions of the can are approximately radius of 6.828cm and height 6.828cm

(b) If x is the width and y the length of the enclosure then the area is A=l*w and the perimeter is 2x+y.

Since the perimeter is 800 we have 2x+y=800 ==> y=800-2x

Then the area is A=x(800-2x). We want to maximize the area so we find the first derivative and set equal to zero:

`(dA)/(dx)=-4x+800` ; -4x+800=0 ==>x=200 and y=400.

Thus the maximum area is `80000m^2` with the dimensions 200m x 400m

Approved by eNotes Editorial Team