A gardener having $120 \ \mathrm m$ of fencing wishes to enclose a rectangular

Question

A gardener having $120 \ \mathrm m$ of fencing wishes to enclose a rectangular plot of land and also erect a fence across the land parallel to two of the sides. Find the maximum area he can enclose.

My attempt:

Let $x$ and $y$ be the length and breadth of rectangle. Then $$2(x+y)=120\implies x+y=60 \implies y=60-x$$

Now, $$A=xy\implies A=x(60-x)\implies A=60x-x^2$$ Then, $A'=60-2x$

$A'=0$ gives $x=30$ and $y=30$. Is it correct?

Answer 1

Not correct. You forgot to take into account the extra fence that goes across the land.

So, you need to start with $2x +3y=120$

Answer 2

No, your answer is not correct.

Your $$2(x+y)=120$$ should have been $$3y+2x=120$$

You forgot about the middle fence.

My final answer is $x=30$ and $y=20$ with the maximum area of $600$