Optimization Problems

Question

A rectangular piece of paper with perimeter 100 cm is to be rolled to forma cylindrical tube. Find the dimensions of the paper that will produce a tube with maximum volume. 

Answer 1

A rectangular piece of paper with perimeter 100 cm is to be rolled to form a cylindrical tube. Find the dimensions of the paper that will produce a tube with maximum volume.

Dimensions of paper: x by y

2x + 2y = 100

x + y = 50

Volume of cylinder: V = Bh = pi r^2 h

Let x/2 = r and y = h = 50 – x:

V = pi (x/2)^2 (50 – x)
Domain: 0 < x < 50

V = pi/4 x^2 (50 – x)

Extrema when V’ = 0

V’ = pi/4 (2x(50 – x) – x^2) = 0

100x – 2x^2 – x^2 = 0

100x – 3x^2 = 0

x(100 – 3x) = 0
Domain excludes x = 0

100 – 3x = 0

x = 100/3 = 33+1/3 cm
y = 50 – 33 – 1/3 = 16+2/3 cm

Answer 2

Let w and l be the dimensions of the paper. We know that 2(l+w)=100  or l+w=50.

 

Volume of a cylinder is given by:

 

V=pi*r²*h.

 

First case:

h=l; then r=w/(2*pi) and V=w²l/(4*pi). Using l+w=50 we obtain:

 

V=w²(50-w)/(4*pi); 

V'(w)=(100w-3w²)/(4*pi); V'(w)=0 at w=0 and w=100/3. Obviously, w=100/3 cm gives maximum volume. l=50-100/3=50/3 cm.

V(100/3)=10000/9*(50-100/3)/(4*pi)=125000/(27*pi)

 

Second case h=w, then the same considerations are applied and the same result is obtained for volume.

 

Answer: 100/3 cm and 50/3 cm; V_max=125000/(27*pi)