# Inequalities about area and perimeter

"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,
$$l>w$$ $$2(l+b)<40$$ $$lb>40$$

• I need help please – Hshs Sbvv Jan 21 at 14:21
• How would you write the specifications in mathematical terms, using symbols? You can set width$=w$ and length$=l$. – Matti P. Jan 21 at 14:21
• We have that: $$l \lt w$$ $$lw \ge 40$$ $$2(l+w) \le 40$$ Can you continue? – Mohammad Zuhair Khan Jan 21 at 14:25
• The area must be greater than 40$cm^2$, right? – Andreas Jan 21 at 15:06

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 $$0 and substituting in the inequality for the area we get $$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 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 $$(10-2\sqrt{15}, 10 )$$