Part a)
To begin with this question we must first find the formulae necessary to work with. We are dealing with surface area (SA) and volume of a cylinder.
SA = 2pi(r)h + 2pi(r^2)
V = pi(r^2)h
We are saying r = x, and we must solve for x when the volume of the can is a maximum. We also know that the SA is equal to 88 cm^2. The question states that the SA value given also includes a lid that overlaps 1/2 cm over the edge, so we need to add an extra 2pi(r)h where h = 1/2 to our formula. We can use the corrected formula to solve for h in terms of x. We are also told to use pi = 22/7.
SA = 2pi(x)h + 2pi(x^2) + pi(x) = 88
2pi(x)h = 88 - 2pi(x^2) - pi(x)
h = (88 - 2pi(x^2) - pi(x))/(2pi(x))
h = 44/(pi(x)) - x - 1/2
Now that we have a formula for h, we can insert this into the Volume formula.
V = pi(x^2)h
V = pi(x^2)(44/pi(x) - x - 1/2)
V = 44x - pi(x^3) - (pi/2)(x^2)
To determine the maximum volume, we must take the derivative of the volume and determine the values of x when dV/dx = 0.
dV/dx = 44 - 3pi(r^2) - pi(x) = 0
-3pi(r^2) - pi(x) + 44 = 0
a = -3pi
b = -pi
c = 44
Now we can use the quadratic formula to solve for x.
x = (-b +/- sqr(b^2 - 4ac))/2a
x = (pi +/- sqr(pi^2 -4(-3pi)(44)))/(-6pi)
x = (22/7 +/- 286/7)/(-132/7)
x = -7/3 and x = 2
Since we cannot use a negative integer, the radius must be equal to 2 cm at the maximum.
Part b)
Now that we have shown the radius is 2 cm at the maximum volume. We can solve for h and solve for the maximum volume.
h = 44/(pi(x)) - x - 1/2
h = 44/(pi(2)) - 2 - 1/2
h = (44 - 5pi)/(2pi)
V = pi(r^2)h
V = pi(2^2)(44 - 5pi)/(2pi)
V = 88 - 10pi
V = 88 - 10(22/7)
V = 396/7
V = 56 4/7 cm^3
The maximum volume using the radius of 2 cm gives a volume of 56 4/7 cm^3.
