Inequalities about area and perimeter

Question

"A gardener is laying out a rectangular lawn. His specifications are that the area $(A)$ must be greater than $40$cm but the perimeter $(P)$ must be less than $40$cm. if the width of the lawn $(w)$ has to be less than the length$(l)$, find the range of possible values for the width of the lawn"
Or equivalently,
$$l>w$$ $$2(l+b)<40$$ $$lb>40$$

Answer 1

Set width $=w$ and length $=l$. Now the area is $w \cdot l$ hence you want $w \cdot l \geq 40$ and the perimeter is $2(w+l)$, thus $2(w+l)\leq 40$ or $w+l \leq 20$. Furthermore $w,l >0$ because it's a geometric problem and $w \leq l$ by hypothesis. Combining two inequalities we also get $0<w \leq l \leq 20 -w$ and substituting in the inequality for the area we get $40 \leq w \cdot l \leq w(20-w)=20w-w^2$ i.e. $w^2-20w+40 \leq 0$
Now $w_{1,2}=10 \pm \sqrt{100-40}=10 \pm \sqrt{60}=10 \pm 2 \sqrt{15} $ hence $w \in (10-2\sqrt{15},10+2\sqrt{15})$ but from $0<w \leq l \leq 20 -w$ we have $w=l=20-w$ when $w=20-w$ i.e. $w=10$, so we must have $w \leq 10$. Thus the range of $w$ is $(10-2\sqrt{15}, 10 )$