Question 1082325
  *[illustration de4.png].
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So you know that,
{{{2X^2+2(2X)Z+2XZ=300}}}
{{{2X^2+4XZ+2XZ=300}}}
{{{2X^2+6XZ=300}}}
{{{X^2+3XZ=150}}}
{{{Z=(150-X^2)/(3X)}}}
The volume of the box is,
{{{V=2X(X)(Z)}}}
{{{V=2X^2((150-X^2)/(3X))}}}
{{{V=(2/3)X(150-X^2)}}}
You can maximize by taking the derivative and setting it equal to zero,
{{{dV/dX=(2/3)X(-2X)+(2/3)(150-X^2)}}}
{{{dV/dX=-2(X^2-50)}}}
So,
{{{X^2-50=0}}}
{{{X^2=50}}}
{{{X=sqrt(50)}}}
{{{X=5sqrt(2)}}}
So then,
{{{2X=10sqrt(2)}}}
and
{{{Z=100/(3sqrt(50))}}}
{{{Z=(10/3)sqrt(2)}}}