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GATE CS 2012 · General Aptitude

Question 62

MCQPrinted as GA Q.622 marksWrong answer: −2/3Easy (estimate)

Question

A political party orders an arch for the entrance to the ground in which the annual convention is being held. The profile of the arch follows the equation y=2x−0.1x2y = 2x - 0.1x^2 where yy is the height of the arch in meters. The maximum possible height of the arch is

Choose one option
Official answer and solution
Official answer
B
From
Official GATE 2012 answer key

Solution

Find the maximum value of the quadratic function y = 2x - 0.1x^2.

  1. The equation for the height profile is y=2x−0.1x2y = 2x - 0.1x^2.

  2. Find the stationary point:

dydx=2−0.2x=0  ⟹  x=20.2=10\frac{dy}{dx} = 2 - 0.2x = 0 \implies x = \frac{2}{0.2} = 10
  1. Evaluate maximum height at x=10x = 10:
y=2(10)−0.1(10)2=20−10=10 metersy = 2(10) - 0.1(10)^2 = 20 - 10 = 10\text{ meters}
  1. Thus, the maximum possible height is 10 meters (Option B).
Final answer

B

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper
GATE 2012 Computer Science and Information Technology (CS)
Question
62 of 65, printed as GA Q.62
Section
General Aptitude

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