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Question Number 5414 by FilupSmith last updated on 14/May/16

How do we prove the area of a square/rectangle is at its maximum when both sides are equal. So, if we have a rectangle of sides a and b, how do we show its maximized when a=b?

$$\mathrm{How}\:\mathrm{do}\:\mathrm{we}\:\mathrm{prove}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{square}/\mathrm{rectangle} \\ $$$$\mathrm{is}\:\mathrm{at}\:\mathrm{its}\:\mathrm{maximum}\:\mathrm{when}\:\mathrm{both}\:\mathrm{sides}\:\mathrm{are}\:\mathrm{equal}. \\ $$$$ \\ $$$$\mathrm{So},\:\mathrm{if}\:\mathrm{we}\:\mathrm{have}\:\mathrm{a}\:\mathrm{rectangle}\:\mathrm{of}\:\mathrm{sides}\:{a}\:\mathrm{and}\:{b}, \\ $$$$\mathrm{how}\:\mathrm{do}\:\mathrm{we}\:\mathrm{show}\:\mathrm{its}\:\mathrm{maximized}\:\mathrm{when}\:{a}={b}? \\ $$

Answered by 123456 last updated on 14/May/16

lets a retangle with side a and b lets s be area and p the perimeter p=2a+2b s=ab lets get all rectangle with constant perimeter, we have a=(p/2)−b 0≤b≤p/2 s=ab =((p/2)−b)b =−b^2 +((pb)/2) finding critic point s_b =−2b+(p/2) s_b =0⇔b=(p/4)⇔a=(p/4) s_(bb) =−2<0 so b=2 is a point of max so “in the conjunt of rectangle with same perimeter, the square have the greatest area”

$$\mathrm{lets}\:\mathrm{a}\:\mathrm{retangle}\:\mathrm{with}\:\mathrm{side}\:{a}\:\mathrm{and}\:{b} \\ $$$$\mathrm{lets}\:{s}\:\mathrm{be}\:\mathrm{area}\:\mathrm{and}\:{p}\:\mathrm{the}\:\mathrm{perimeter} \\ $$$${p}=\mathrm{2}{a}+\mathrm{2}{b} \\ $$$${s}={ab} \\ $$$$\mathrm{lets}\:\mathrm{get}\:\mathrm{all}\:\mathrm{rectangle}\:\mathrm{with}\:\mathrm{constant} \\ $$$$\mathrm{perimeter},\:\mathrm{we}\:\mathrm{have} \\ $$$${a}=\frac{{p}}{\mathrm{2}}−{b}\:\:\:\:\:\mathrm{0}\leqslant{b}\leqslant{p}/\mathrm{2} \\ $$$${s}={ab} \\ $$$$=\left(\frac{{p}}{\mathrm{2}}−{b}\right){b} \\ $$$$=−{b}^{\mathrm{2}} +\frac{{pb}}{\mathrm{2}} \\ $$$$\mathrm{finding}\:\mathrm{critic}\:\mathrm{point} \\ $$$${s}_{{b}} =−\mathrm{2}{b}+\frac{{p}}{\mathrm{2}} \\ $$$${s}_{{b}} =\mathrm{0}\Leftrightarrow{b}=\frac{{p}}{\mathrm{4}}\Leftrightarrow{a}=\frac{{p}}{\mathrm{4}} \\ $$$${s}_{{bb}} =−\mathrm{2}<\mathrm{0} \\ $$$$\mathrm{so}\:{b}=\mathrm{2}\:\mathrm{is}\:\mathrm{a}\:\mathrm{point}\:\mathrm{of}\:\mathrm{max} \\ $$$$\mathrm{so} \\ $$$$``\mathrm{in}\:\mathrm{the}\:\mathrm{conjunt}\:\mathrm{of}\:\mathrm{rectangle}\:\mathrm{with}\:\boldsymbol{\mathrm{same}} \\ $$$$\boldsymbol{\mathrm{perimeter}},\:\mathrm{the}\:\mathrm{square}\:\mathrm{have}\:\mathrm{the}\:\mathrm{greatest} \\ $$$$\mathrm{area}'' \\ $$

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