$($a$)$ Consider the flollowing statement.

or all the rectangles with a glven area, the one with smallest perimeter is a square.

The following is a proposed proof for the given statement.

Let the rectangle have slides $x$ and $y$ and area $A,50A=xy$ or $y=\frac{A}{x},$

The problem is to maximize the perimeter;

$f(x)=2x+2y=2x+\frac{2x}{x}$

Now $P^{\text{'}}(x)=2-\frac{24}{x^2}=\frac{2(x^2-A)}{x^2}$

So the critical number is $x=A.$

Since $P^{\text{'}}(x)<mn>0</mn>$ for $P^{\text{'}}(x)>0x>\sqrt[\phantom{2}]{A}x=\sqrt[\phantom{2}]{A}\sqrt[\phantom{2}]{7}\frac{A}{\sqrt[\phantom{2}]{A}}=\sqrt[\phantom{2}]{A}{P}^{\text{'}}(x)P^{\text{'}}(x)=2-\frac{24}{x}=\frac{2(x-A)}{x}x=\sqrt[\phantom{2}]{A}x=A\frac{A}{\sqrt[\phantom{2}]{A}}=A\frac{A}{\sqrt[\phantom{2}]{A}}=\sqrt[\phantom{2}]{A}0$ and $P^{\text{'}}(x)>0$ for $x>\sqrt[\phantom{2}]{A},$ there $isanabsolute$ minimum $atx=\sqrt[\phantom{2}]{A}.$

Thurtur $of$ tha receingte and $\sqrt[\phantom{2}]{7}$ and $\frac{A}{\sqrt[\phantom{2}]{A}}=\sqrt[\phantom{2}]{A},so$ the receangle $is$ a square.

Idently the error$(s)in$ the proposed proof. $(Select$ all that apply.$)$

The second sentence should say minimire instead $of$ maximise.

$In$ the third sentence, the derivative $P^{\text{'}}(x)is$ incorrect; $it$ should $be$

$P^{\text{'}}(x)=2-\frac{24}{x}=\frac{2(x-A)}{x}$

$In$ the fourth sentence, the ortical number $is$ incorrect; $it$ should $bex=\sqrt[\phantom{2}]{A}$ instead $ofx=A.$

The finis sentence should save $absolute$ maximum instead $of$ aboolute minimum.

The sixth sentence has $an$ incorrect colculation; $it$ should say $\frac{A}{\sqrt[\phantom{2}]{A}}=$A instead $of$

$\frac{A}{\sqrt[\phantom{2}]{A}}=\sqrt[\phantom{2}]{A}$