The design of a digital box camera maximizes the volume while keeping the sum of the dimensions at 6 inches. If the length must be 1.5 times the height, what should each dimension be?
Hint: Let x represent one of the dimensions, then define the other dimensions in terms of x.
Help?

Question

The design of a digital box camera maximizes the volume while keeping the sum of the dimensions at 6 inches.  If the length must be 1.5 times the height, what should each dimension be? 
 Hint: Let x represent one of the dimensions, then define the other dimensions in terms of x. 
 Help?

anonymous · Answer

I am assuming that it is a 2D shape. Is that right?

anonymous · Answer

Let the perimeter equal P

P = 2l+2w
6 = 2*(1.5x)+2x
6 = 3x+2x
6 = 5x
x = 6/5
x = 1.2 inches

Width = 1.2 inches
Height = 1.5*width = 1.8 inches

anonymous · Answer

thx!

anonymous · Answer

could you help me with another?

directrix · Answer

From Text-a-Friend @ridley9510 

  L = 1.5 H = (3/2) H 
 (3/2)H + H + W = 6 so 
 W = 6 - H - (3/2)H = 6 - (5/2) H 

 V = L*H*W = (3/2) H * H * (6 - (5/2) H) 

 volume is now a function of H alone and we can maximize it by 
 taking the derivative. To make it easier, replace H by x. 

 V = (3/2) x * x * (6 - (5/2) x) = 9 x^2 - (15/4) x^3 
 dV/dx = 18 x - (45/4) x^2 = 0 
 this has the solution x = 0 
 obtained by factoring out x. 
 Then you have 18 - (45/4)x = 0 
 x = 72/45 as the other root. 

 You can calculate the inflection points or simply graph out the function to be sure whether 
 the points are minima or maxima. In this case, H = 72/45 = 1.6 is a maximum giving for the 
 dimensions: 
 H = 1.6 
 L = 2.4 
 W = 2.0 

 and the volume V = 7.68