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**