A box with no top must have a base twice as long as it is wide, and the total surface area of the box is to be 54 meters squared. What is the maximum possible volume of such a box?
I have the work. I just want to make sure I'm doing everything right here.
\(\displaystyle \
\L\
\begin{array}{l}
SA = 54m^2 \\
V = 2x^2 y \\
54 = 2x^2 + 4xy \\
y = \frac{{54 - 2x^2 }}{{4x}} \\
\to V = 2x^2 \left( {\frac{{54 - 2x^2 }}{{4x}}} \right) \\
V = \frac{{108x^2 - 4x^4 }}{{4x}} \\
V = 27x - x^3 \\
\to \frac{{dV}}{{dx}} = 27 - 3x^2 \\
0 = 3(9 - x^2 ) \\
0 = 3(3 - x)(3 + x) \to \pm 3{\rm but not } - 3 \\
y = \frac{{54 - 2(3)^2 }}{{4(3)}} \to y = 3 \\
\to V = 2(3)^2 (3) \\
V = 54m^3 \\
{\rm Therfore, the maximum volume is 54m}^{\rm 3} {\rm and occures when} \\
{\rm x = 3 and y = 3}{\rm .} \\
\end{array}
\\)
So, how does it look?
I have the work. I just want to make sure I'm doing everything right here.
\(\displaystyle \
\L\
\begin{array}{l}
SA = 54m^2 \\
V = 2x^2 y \\
54 = 2x^2 + 4xy \\
y = \frac{{54 - 2x^2 }}{{4x}} \\
\to V = 2x^2 \left( {\frac{{54 - 2x^2 }}{{4x}}} \right) \\
V = \frac{{108x^2 - 4x^4 }}{{4x}} \\
V = 27x - x^3 \\
\to \frac{{dV}}{{dx}} = 27 - 3x^2 \\
0 = 3(9 - x^2 ) \\
0 = 3(3 - x)(3 + x) \to \pm 3{\rm but not } - 3 \\
y = \frac{{54 - 2(3)^2 }}{{4(3)}} \to y = 3 \\
\to V = 2(3)^2 (3) \\
V = 54m^3 \\
{\rm Therfore, the maximum volume is 54m}^{\rm 3} {\rm and occures when} \\
{\rm x = 3 and y = 3}{\rm .} \\
\end{array}
\\)
So, how does it look?