Question 1094819
Bottom dimensions x and x.
height y.

{{{y=x+2}}},  ;
{{{y*x^2=96}}}
and just those should be enough without using the surface area description {{{x^2+4xy=112}}}.


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{{{(x+2)*x^2=96}}}
{{{x^3+2x^2-96=0}}}
One of the solutions is x=4.

This would make  {{{highlight(system(x=4,y=6))}}}




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check
6*4*4=96?
24*4=96
96
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4^2+4*4*6=112?
16+4*24=112
16+96
112
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Both conditions work.