SL 5.8—Testing for max and min, optimisation. Points of inflexion

The perimeter of a rectangle with length $x$ and width $y$ is $2x+2y=60$, so $x+y=30$, giving $y=30-x$.

Area $=x\times y=x(30-x)$.

The area in terms of $x$ alone is $A=x(30-x)$, which matches the first option.

We need $A$ as a function of $x$ only, using the perimeter constraint $2x+2y=60$.

This would require $y=60-x$, which comes from $x+y=60$, i.e., a perimeter of $2(x+y)=120\ne 60$. Eliminated.

This would give $y=15-x$, so $x+y=15$ and perimeter $=30\ne 60$. Eliminated.

This doubles the area expression without justification — $A=xy$, not $2xy$. Eliminated.

From $2x+2y=60\Rightarrow y=30-x$, so $A=x(30-x)$ is correct.

With perimeter $2x+2y=60$, we get $y=30-x$, so $A(x)=x(30-x)=30x-x^2$.

$A^{\prime }(x)=30-2x=0\Rightarrow x=15$

$A^{\prime \prime }(x)=-2<0$, confirming $x=15$ gives a local maximum.

The area is maximised when $x=15$ metres (a square enclosure).

Maximise $A=x(30-x)$ over $0<x<30$ by finding the critical point.

At $x=30$, the width $y=0$, giving zero area. This is a boundary minimum, not a maximum.

At $x=20$: $A=20\times 10=200$ m². But we can check $x=15$: $A=15\times 15=225$ m², which is larger.

At $x=10$: $A=10\times 20=200$ m², same as $x=20$ by symmetry, less than 225 m².

$A^{\prime }(x)=30-2x=0$ gives $x=15$, with maximum area $225$ m².