Question 1142709
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When the four squares at the corners are cut and the sides folded up, the base of the open box has 
dimensions  (16-2x) cm  and  (30-2x) cm,  while its height is x cm.



So, the volume of the open box is


    V(x) = x*(16-2x)*(30-2x)


    V(x) = x*(4x^2 - 92x + 480) = 4x^3 - 92x^2 + 480x.


To find the maximum volume, first take the derivative of the volume over x and then equate it to zero:


    V'(x) = 12x^2 - 184x + 480 = 0.


Simplify the equation


    3x^2 - 46x + 120 = 0.


Find the roots using the quadratic formula


{{{x[1,2]}}} = {{{(46 +- sqrt(46^2 - 4*3*120))/(2*3)}}} = {{{(46 +- 26)/6}}}.


The root  {{{x[1]}}} = {{{(46 + 26)/6}}} = 12  is not the solution to the problem, since (16-2x) is NEGATIVE.


{{{x[2]}}} = {{{(46 - 26)/6}}} = {{{20/6}}} = {{{3}}}{{{2/3}}} cm  is the solution to the problem.


<U>ANSWER</U>.  Optimum dimension of the squares to maximize the volume is  {{{3}}}{{{2/3}}} cm.


See the plot below of the volume V(x) as a function of x.



    {{{graph( 500, 500, -10, 20, -800, 800,
          x*(16-2x)*(30-2x)
)}}}


        Plot  V(x) = {{{x*(16-2x)*(30-2x)}}}

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