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GATE CS 2012 Q62 — General Aptitude

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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
 

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Answer: B

To find the maximum possible height of the arch, we need to find the maximum value of the quadratic function:

\ny=2x−0.1x2y = 2x - 0.1x^2\n\n

Method 1: Using Calculus

\n

Differentiate yy with respect to xx to find the critical point:

\ndydx=ddx(2x−0.1x2)=2−0.2x\frac{dy}{dx} = \frac{d}{dx}(2x - 0.1x^2) = 2 - 0.2x\n\n

Set the first derivative to zero to find the extremum:

\n2−0.2x=0  ⟹  0.2x=2  ⟹  x=102 - 0.2x = 0 \implies 0.2x = 2 \implies x = 10\n\n

To verify that this point is a maximum, we check the second derivative:

\nd2ydx2=−0.2<0\frac{d^2y}{dx^2} = -0.2 < 0\n

Since the second derivative is negative, x=10x = 10 corresponds to a local maximum.

\n\n

Substitute x=10x = 10 back into the original equation to find the maximum height:

\nymax=2(10)−0.1(10)2=20−10=10 metersy_{\text{max}} = 2(10) - 0.1(10)^2 = 20 - 10 = 10\text{ meters}

\n\n

Method 2: Completing the Square

\n

We can rewrite the quadratic equation by completing the square:

\ny=−0.1(x2−20x)y = -0.1(x^2 - 20x)\ny=−0.1(x2−20x+100)+10y = -0.1(x^2 - 20x + 100) + 10\ny=−0.1(x−10)2+10y = -0.1(x - 10)^2 + 10\n\n

Since the term −0.1(x−10)2-0.1(x - 10)^2 is always non-positive (i.e., ≤0\le 0), the maximum value of yy occurs when this term is 00, which happens at x=10x = 10. Thus, the maximum height is 1010 meters.

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