Question 1143072
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The dimensions of the box are  (12-2x) inches (each the length and the width of the base).


The height of the box is x inches.


The volume of the box is  

    V(x) = x*(12-2x)*(12-2x) = x*(144 - 48x + 4x^2) = 4x^3 - 48x^2 + 144x.


To determine the maximum of the function V(x), take the derivative

    V'(x) = 12x^2 - 96x + 144


and equate it to zero

    12x^2 - 96x + 144 = 0.


It gives

    x^2 -8x + 12 = 0


Factor left side

    (x-6)*(x-2) = 0.


The two roots are  x= 6  and  x= 2.


The volume at x= 6 is equal to zero, so this root, although provides the local minimum, does not give a real solution.


The volume at x= 2 is   V(2) = 2*(12-2*2)*(12-2*2) = 2*8*8 = 128 cubic inches.


It is the real maximum and the real solution to the problem.



    {{{graph( 330, 330, -2, 10, -20, 200,
          4x^3 -48x^2 + 144x
)}}}


    Plot  V(x) = 4x^3 -48x^2 + 144x.
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