Sigma Percentile
JEE Advanced 1981
LEVELJEE Main

Animated Solution for Mathematics - Differentiation: Let and be two real variables such that and . Find the minimum value of .

Enter Numerical Value:

Visualized Solution

Visualizing the Constraint

  • Given: and .
  • The constraint represents a rectangular hyperbola in the first quadrant.
  • We need to find the minimum value of the sum .

Expressing as a Function of

  • From the constraint , we can write as a function of :
  • This substitution is valid since .

Defining the Sum Function

  • Let the sum be .
  • Substituting into the equation:

Finding the First Derivative

  • To find the extrema, we differentiate with respect to :

Solving for Critical Points

  • Set to find critical points:
  • (since )

The Second Derivative Test

  • Differentiate again to find :
  • At , .
  • Since the second derivative is positive, is a point of local minimum.

Calculating the Minimum Value

  • Substitute back into the sum function:
  • The minimum value of is 2.

Alternative Method: AM-GM Inequality

  • Alternative Method (AM-GM Inequality):
  • Equality holds when .

The Sigma Insight: Maxima and Minima

Solution Diagram

Analyzing the Setup

We are operating in the first quadrant of the Cartesian plane, constrained by the rectangular hyperbola defined by the equation:
We aim to minimize the sum . Geometrically, this is equivalent to finding the smallest constant such that the line intersects the hyperbola.

The Calculus Path

Bridging Variables
Since and , we can express as . This allows us to define the sum as a single-variable function:
To find the critical points, we differentiate with respect to :
Setting the derivative to zero to find the stationary point:
Given the constraint , we discard the negative root and conclude that the critical point occurs at .

The Second Derivative Test

Confirming the Minimum
To verify that this critical point is a minimum, we examine the second derivative:
Evaluating this at our critical point :
Since the second derivative is positive, the function is concave up at , confirming a local minimum. Substituting back into the sum, we find:

The Elegant Shortcut

AM-GM Inequality
For any positive real numbers and , the Arithmetic Mean-Geometric Mean (AM-GM) inequality states:
Substituting our constraint into the inequality:
This confirms that the minimum value of the sum is 2, occurring precisely when .

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