Question 39

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Jambmaths question: 

Find the dimension of the rectangle of greatest areas which has a fixed perimeter p.

Option A: 

square of sides p

Option B: 

square of sides 2p

Option C: 

square of sides p/2

Option D: 

square of  p/4

Jamb Maths Solution: 

$\begin{align}  & A=l\times b----(i) \\ & p=2(l+b)---(ii) \\ & p=2l+2b \\ & l=\frac{p-2b}{2}----(iii) \\ & \text{substitute }l=\frac{p-2b}{2}\text{ into }(i) \\ & A=\frac{p-2b}{2}\times b \\ & A=\frac{pb-2{{b}^{2}}}{2} \\ & \frac{dA}{db}=\frac{p-4b}{2} \\ & A\text{ will be maximum when }\frac{dA}{db}=0 \\ & \frac{p-4b}{2}=0 \\ & p=4b------(iv) \\ & \text{Since }A=\frac{p-2b}{2}\times b \\ & A=\frac{4b-2b}{2}\times b={{b}^{2}} \\ & \text{since }p=4b,\text{  }b=\frac{p}{4} \\ & A={{\left( \frac{p}{4} \right)}^{2}}=\frac{{{p}^{2}}}{16} \\\end{align}$

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