Raphael has a art sheet of $2m$ by $3m$. In order to frame an open box, he cuts four equal squares from the corners of the sheet, folds it along the cuts and glues the lateral faces of the box along their common edges. If the box is to have the largest possible volume, what should be the length of the sides of the squares he cut ?
My work :
Framing equations,
The height = $x$ m
The length = $(3 – 2x)$ m
The width = $(2 – 2x)$ m
Applying AM-GM inequality gives:
$$(2-2 x)(3-2 x) x \leq\left(\frac{(2-2 x)+(3-2 x)+x}{3}\right)^{3}$$
Now, RHS has a dependency on $x$.
Here is where I am stuck, how to proceed further ? The book instructs to not use derivatives.
Possible duplicate with the same technique: here, however it uses derivatives and the 1st answer doesn't answer my case.
Best Answer
Hint:
In such cases, you should consider taking the means for $3-2x, \alpha(2-2x), \beta x$ (assuming you got those terms correctly), for suitable coefficients $\alpha, \beta$.
These unknown coefficients can be chosen to meet the following desirable conditions to use AM-GM inequality:
(a) The sum of the terms should be a constant. i.e. $-2-2\alpha + \beta = 0$
(b) Equality is possible. i.e. there is an allowable value of $x$. s.t. $3-2x=\alpha(2-2x)=\beta x > 0$.
Solving these, you should get $x = \frac16(5-\sqrt7)$.
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P.S. of course derivative approach would also give you the same $x$.