A box is to be made out of a 12 by 20 piece of cardboard. Squares of equal size will be cut out of each corner, and then the ends and sides will be folded up to form a box with an open top. Find the length L , width W , and height H of the resulting box that maximizes the volume.

Box Dimensions:

L  = 15.15 ul

W = 7.15 ul

h = x = 2.43 ul

V(max) =    263.22 cu

Step-by-step explanation:

We call x the length of the square to be cut in the corners then:

Are of the base of the box is:

(20 - 2*x)  is the future length of the box and

(12 - 2*x) will be the width

The heigh is x  then the volume of the box is:

V = ( 20 - 2*x )* ( 12 - 2*x ) * h

And the volume as a function of x is:

V(x) = ( 20 - 2*x) * ( 12 - 2*x ) * x     or   V(x) = (240 -40*x -24*x + 4*x²) * x

V(x) =  240*x - 64*x² + 4*x³

Taking derivatives on both sides of the equation we get:

V´(x) = 240 - 128*x + 12*x²

V´(x)  =  0              240 - 128*x + 12*x²  = 0    or     60 - 32*x + 3*x²

3*x²  -  32*x  +  60  = 0

Solving:

x₁,₂  =  32  ± √ (32)² - 4*3*60 ]/ 2*3

x₁,₂  =  32  ± √ 1024 - 720 )/6

x₁,₂  = ( 32  ± √ 304 )/6

x₁,₂  = ( 32  ± 17.44 )/6

x₁  =  8.23      ( we dismiss this solution because is not feasible 2*x > 12

x₂ = 2.43  u.l ( units of length)

Then

L  =  20 - 2*x      L  = 20  - 4.85     L  = 15.15 ul

W =  12 - 2*x      W = 12 - 4.85      W = 7.15 ul

h  =  2.43 ul

V = 2.43*7.15*15.15  cubic units

V =  263.22 cu

To see if when x = 2.43   function V has a maximum we go to the second derivative

V´´(x)  = - 128  + (24)*2.43

V´´(x) = - 69.68      as        V´´(x)  < 0   then we have a maximum for V(x) in the point x = 2.43

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c > -10.55

step-by-step explanation: