Question 728575
{{{drawing(400,300,-2,24,-1.5,17.5,
red(rectangle(0,0,3,3)),red(rectangle(19,0,22,3)),
red(rectangle(0,14,3,17)),red(rectangle(19,14,22,17)),
blue(rectangle(3,3,19,14)),rectangle(0,0,22,17),
locate(10,0,22),locate(22.2,9,17),
locate(1.3,14,x),locate(3.2,16,x),
locate(9,14,22-2x),locate(3.2,9,17-2x)
)}}} Cut along red lines; fold along blue lines.
{{{The blue base has a surface area of
{{{(22-2x)(17-2x)}}} square inches.
The box has a height of {{{x}}} inches.
The volume of the box is
{{{V=x(22-2x)(17-2x)}}}cubic inches.
 
The expression above gives you the volume {{{V}}} as a function of {{{x}}} , but the fact that {{{x}}} is the measure of the side of the squares cut off imposes some restrictions:
{{{x>0}}} and {{{17-2x>0}}} --> {{{17>2x}}} --> {{{17/2>x}}} --> {{{x<8.5}}} or {{{x<8&1/2}}}
So the variable {{{x}}} must comply with
{{{0<x<8.5}}}
That is the domain of the function {{{V}}}.
 
{{{V=152.5}}} cubic inches, so we must solve
{{{x(22-2x)(17-2x)=152.5}}} and find one or more solutions with {{{0<x<8.5}}} .
That is a cubic (degree 3) equation, and may be difficult to solve.
There could be a rational solution somewhere that could be used to reduce the equation to a quadratic equation.
Otherwise, if the solutions turn out to be irrational, it is a question of finding approximate solutions by painful calculations. The request to round answers to the nearest hundredth suggests that is the case.
A graphing calculator, or some computer software could help with those calculations. The question about the domain suggest that using a graphing calculator is expected. If you have been using a graphing calculator in class, that must be what is expected.
The polynomial function {{{V=x(22-2x)(17-2x)}}} would be zero at x=0 and x=8.5, and we know that it is positive in between.
Logic tells us that {{{V=x(22-2x)(17-2x)}}} has a maximum in the domain {{{0<x<8.5}}} .
For {{{x=1}}} --> {{{V=1*20*15=300>152.5}}} , so 152.5 is not the maximum.
We should expect two solutions, one on each side of the maximum.
I find that there is a solution between 0.445 and 0.450, which would be {{{highlight(0.45)}}} rounded to the nearest hundredth.
There is another solution between 7.115 ans 7.12 which would round to {{{highlight(7.12)}}}