Math Help please

Question

Min Surface of a box with a volume of 2.4m3

Max volume of a cylinder with a surface area of 10.5m2

Answer 1

2.4 m^3 = volume

cube root of 2.4 = about 1.339 m. for the length of a side of the cube. A cube has minimal surface area for a given volume of a box

surface area = 6 times 2.4^(2/3) = 6(1.339)^2 = about 10.758 m^2

for a cylinder with surface area = 10.5 m^2 = 2pi(r)^2 + 2rhpi =

2pir(r+h) = 10.5

r+h = 10.5/2pir

h = 10.5/2pir -r = 5.25/pir - r

Volume of a cylinder = hpir^2 = V(r) = [(10.5)/2pir - r](pir^2)

V(r) = 5.25r - pir^3

take the derivative and set = 0

V'(r) = 5.25 -3pir^2 = 0

r^2 = 5.25/3pi

r = sqr(5.25/3pi) = about 0.557 = radius of the cylinder

h = 10.5/2pir -r = 1.67/.557 - .557 = 2.443 = height of the cylinder

V= 5.25(.557) - pi(.557)^3 = 2.924 - .543 = 2.381 m^3

max V = 2.381 m^3 for surface area = 10.5 m^2

V=hpir^2 = 2.443(pi).557^2 = 2.381 m^3 = max V