Optimization problem - maximum volume and maximum face area of a parallelepiped.

Question

I've been working on this optimization question for a while now and just can't get it figured out. Here's the information that the question provides...

A 6 m long metal rod is cut into 12 sections to form the edges of a right parallelepiped. Determine the length of the edges of the parallelepiped so that :

a) the volume of the parallelepiped is maximal.

b) the totale face area of the parallelepiped is maximum.

Length = 2x

Width = x

Height = y

Help?

Answer 1

a) Volume V = 2x·x·y = 2x²y and 4·2x + 4x + 4y = 6 or 12x + 4y = 6; y = 1.5 - 3x;

V = 2x²(1.5 - 3x) = 3x² - 6x³; V' = 6x - 18x² = 0; x = 0 or x = 1/3 m. Now V'' = - 36x and V''(1/3) = - 12 < 0

So by second derivative test at x = 1/3 m we have maximum volume: 2x = 2/3 m, x = 1/3 m, y = 0.5 m

V_max = 2/3·1/3·1/2 = 1/9 m³ .

b) Total surface are is S = 2(2x²) + 2(2xy) + 2(xy) = 4x² + 6xy and 12x + 4y = 6, y = 1.5 - 3x;

S = 4x² + 6x(1.5 - 3x) = 4x² + 9x - 18x² = 9x - 14x² ; S' = 9 - 28x = 0, x = 9/28 m

S'' = - 28 = 0; S'' at x = 9/28 m = - 28 < 0. By 2nd derivative test we have minimum: x = 9/28 m, 2x = 9/14m,

y = 3/2 - 27/28 = 15/28 m and S_min = 4(9/28)² + 6(9/28)(15/28) = 1134/784 = 81/56 m² = 1.4464 m².