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CAT 2022 Slot 3 — QA Question 12

Inequality Maximization / MinimizationEasy

If c = 16xy + 49yx for some non-zero real numbers x and y, then c cannot take the value

Answer & solution

  • A

    -60

  • B

    -70

  • C

    60

  • -50

Solution

Easy

Substitute a=xya=\dfrac{x}{y} so c=16a+49ac=16a+\dfrac{49}{a}. By AM–GM the value of such an expression is bounded away from 00: for a>0a>0 it is at least 5656, and for a<0a<0 it is at most −56-56. The "forbidden" value is whatever lies strictly inside the gap (−56,56)(-56,56).

1

Reduce to one variable. Let a=xya=\dfrac{x}{y}:

c=16xy+49yx=16a+49ac=\frac{16x}{y}+\frac{49y}{x}=16a+\frac{49}{a}
2

Case a>0a>0 — apply AM–GM to the two positive terms:

16a+49a2≥16a⋅49a=784=28⇒ c≥56\begin{aligned} &\frac{16a+\frac{49}{a}}{2}\ge\sqrt{16a\cdot\frac{49}{a}}=\sqrt{784}=28\\ &\Rightarrow\ c\ge 56 \end{aligned}
3

Case a<0a<0. Both terms are negative, so by symmetry the maximum is −56-56:

c=16a+49a≤−56c=16a+\frac{49}{a}\le -56
4

Combine. So c≥56c\ge 56 or c≤−56c\le -56; no value of cc lies in (−56,56)(-56,56). Checking the options: −60,−70-60,-70 (≤−56\le-56) and 6060 (≥56\ge56) are all attainable, but −50-50 lies inside the forbidden gap.

cc cannot equal −50\mathbf{-50}.

The reachable values are everything with ∣c∣≥216⋅49=56|c|\ge 2\sqrt{16\cdot49}=56. Just scan the options for the one with ∣c∣<56|c|<56 — only −50-50 qualifies.

CAT 2022 Slot 3 QA Q12: If c = 16 x y + 49 y x for some non-zero real numbers x and y, then c cannot take the value — Solution | TheCATExam