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CAT 2017 Slot 2 — QA Question 19

Basics of Mensuration/PrismEasy

If three sides of a rectangular park have a total length 400 ft, then the area of the park is maximum when the length (in ft) of its longer side is

Answer & solution

Correct answer: 200

Answer: 200

Solution

Easy

Only three sides are fenced: two widths and one length. With the perimeter constraint fixed, the area is a downward parabola in one variable; maximise it (or use AM–GM). The longer dimension turns out to be the single fenced length.

1

Set up. Let the two equal fenced sides be xx and the third side be yy. Total fenced length is 400400.

2x+y=400(three sides)Area A=xy\begin{aligned} &2x + y = 400 \quad\text{(three sides)}\\ &\text{Area } A = xy \end{aligned}
2

Maximise. Substitute y=400−2xy = 400 - 2x and find the vertex of the parabola A(x)A(x).

A=x(400−2x)=400x−2x2(from step 1)⇒ dAdx=400−4x=0⇒ x=100\begin{aligned} &A = x(400 - 2x) = 400x - 2x^2 \quad\text{(from step 1)}\\ &\Rightarrow\ \tfrac{dA}{dx} = 400 - 4x = 0\\ &\Rightarrow\ x = 100 \end{aligned}
3

Longer side. Back-substitute to get yy, the single fenced side.

y=400−2(100)=200(from step 2)⇒ y=200>x=100(so y is the longer side)\begin{aligned} &y = 400 - 2(100) = 200 \quad\text{(from step 2)}\\ &\Rightarrow\ y = 200 > x = 100 \quad\text{(so } y \text{ is the longer side)} \end{aligned}

By AM–GM the product (2x)(y)(2x)(y) is largest when 2x=y2x = y; with 2x+y=4002x+y=400 this gives y=200y=200 directly.

Longer side=200 ft\text{Longer side} = 200 \text{ ft}

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CAT 2017 Slot 2 QA Q19: If three sides of a rectangular park have a total length 400 ft, then the area of the park is maximum when the — Solution | TheCATExam