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

  • $\begingroup$ I need help please $\endgroup$ – Hshs Sbvv Jan 21 at 14:21
  • $\begingroup$ How would you write the specifications in mathematical terms, using symbols? You can set width$=w$ and length$=l$. $\endgroup$ – Matti P. Jan 21 at 14:21
  • $\begingroup$ We have that: $$l \lt w$$ $$lw \ge 40$$ $$2(l+w) \le 40$$ Can you continue? $\endgroup$ – Mohammad Zuhair Khan Jan 21 at 14:25
  • $\begingroup$ The area must be greater than 40$cm^2$, right? $\endgroup$ – 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 $0<w \leq l \leq 20 -w$ 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 $(10-2\sqrt{15}, 10 )$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.