Answer:
A. 14x14x28
B. The maximum volume is 5488 cuibic inches
Step-by-step explanation:
The problem states that the box has square ends, so you can express volume with:
[tex]v=x^{2} y[/tex]
Using the restriction stated in the problem to get another equation you can substitute in the one above:
[tex]4x+y=84\\\\[/tex]
Substituting y whit this equation gives:
[tex]v=x^{2} (84-4x)\\\\v=84x^{2} -4x^{3}[/tex]
Now find the limit of x:
[tex]\frac{84x^{2}-4x^{3}}{dx}=168x-12x^{2}\\\\x=\frac{168}{12}=14[/tex]
Find the length:
[tex]y=84-4(14)=28[/tex]
You can now calculate the maximum volume:
[tex]v=(14)^{2}(28)= 5488[/tex]
