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CAT 2021 Slot 3 — QA Question 5

Number TheoryEasy

The cost of fencing a rectangular plot is ₹ 200 per ft along one side, and ₹ 100 per ft along the three other sides. If the area of the rectangular plot is 60000 sq. ft, then the lowest possible cost of fencing all four sides, in INR, is 

Answer & solution

  • A

    160000

  • 120000

  • C

    100000

  • D

    90000

Solution

Easy

One length costs the premium rate, the opposite length and both widths cost the cheaper rate. Write the cost as a multiple of (3l+2b)(3l+2b) subject to l b=60000l\,b=60000. Minimise the linear expression 3l+2b3l+2b under a fixed product using the AM–GM inequality.

1

Express the fencing cost. Let the sides be ll and bb ft with l b=60000l\,b=60000. One ll-side costs ₹200\textsf{₹}200/ft; the other ll-side and the two bb-sides cost ₹100\textsf{₹}100/ft.

Cost=200 l+100 (l+2b)⇒ Cost=100(3l+2b)\begin{aligned} &\text{Cost}=200\,l+100\,(l+2b)\\ &\Rightarrow\ \text{Cost}=100(3l+2b) \end{aligned}
2

Minimise 3l+2b3l+2b by AM–GM. The arithmetic mean of two positive quantities is at least their geometric mean, with the product 3l⋅2b=6 lb3l\cdot 2b=6\,lb fixed.

3l+2b2≥3l⋅2b(AM≥GM)⇒ 3l+2b≥26 lb=26×60000(lb=60000)⇒ 3l+2b≥2360000=2(600)=1200\begin{aligned} &\frac{3l+2b}{2}\ge\sqrt{3l\cdot 2b} \quad\text{(AM}\ge\text{GM)}\\ &\Rightarrow\ 3l+2b\ge 2\sqrt{6\,lb}=2\sqrt{6\times 60000} \quad\text{(}lb=60000\text{)}\\ &\Rightarrow\ 3l+2b\ge 2\sqrt{360000}=2(600)=1200 \end{aligned}
3

Minimum cost. Equality (the cheapest case) holds when 3l=2b3l=2b, which is feasible alongside lb=60000lb=60000, so the bound is attained.

Costmin⁡=100×1200=120000\begin{aligned} &\text{Cost}_{\min}=100\times 1200=120000 \end{aligned}
Lowest cost=₹ 120000\text{Lowest cost}=\textsf{₹}\,120000
CAT 2021 Slot 3 QA Q5: The cost of fencing a rectangular plot is ₹ 200 per ft along one side, and ₹ 100 per ft along the three other — Solution | TheCATExam