Question 1051078
{{{x=8-y^2}}}
Substitute into q equation,
{{{q=5(8-y^2)y^2}}}
Now q is a function of one variable, y.
To find the maximum, differentiate with respect to y and set the derivative equal to zero.
{{{dq/dy=5(8-y^2)(2y)+5(y^2)(-2y)}}}
{{{dq/dy=80y-10y^3-10y^3}}}
{{{dq/dy=80y-20y^3}}}
So,
{{{80y=20y^3}}}
{{{y^2=4}}}
Remember y is positive,
{{{y=2}}}
So then,
{{{x+2^2=8}}}
{{{x+4=8}}}
{{{x=4}}}
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{{{q[max]=5(4)(4)=80}}}