A service requires the dimensions of a rectangle box are such that the length plus twice the width plus twice the height do not exceed $274cm$ ($l+2w+2h \leq 274$). What is the maximum volume of the box that the service accepts to receive??
That means that we want to maximize $V(l, w, h)=lwh$ subject to $l+2w+2h \leq 274$, right??
To find the critical points we do the following:
$$\nabla V(l, w, h)=\overrightarrow{0} \Rightarrow (wh, lh, lw)=(0, 0, 0)$$
How can we continue to get the critical points??
I do not see the Lagrange multipliers of the tag being used, where is the $\lambda$? It is obviously best to use the full $274$, so we maximize $lwh$ subject to $1+2w+2h=274$.
The Lagrangian is $lwh-\lambda(l+2w+2h)$. Set the partial derivatives equal to $0$. We get $wh-\lambda=0$, $lh-2\lambda=0$, $lw-2\lambda=0$. We have the additional equation $l+2w+2h=274$. Solve.
Alternately, we could note that $l=274-2w-2h$, substitute in the expression for $V$, and solve as a $2$ variable maximization problem.
Amusing fact: When one does the arithmetic, the maximum volume turns out to be about $190470.6$ cm^$3$. The answer given is $11664$ in$^3$!!
When one converts, one finds that the answer given is approximately right, but not quite. Why did this happen? It turns out that US postal rules for priority mail specify that $l+2w+2h$ must not exceed $108$ inches. And when one does the calculation with $108$, it turns out that the maximum volume is indeed $11664$ in$^3$. But $108$ inches is not exactly $274$ cm, it is $274.32$ cm. The problem was partly translated to metric, but not exactly, and the answer remained in cubic inches.