Question 4

Mathematics Basic Algebra Hard

Suppose that the temperature at a point (x,y), on a metal plate is \(T(x,y)=4x^2-4xy+y^2\), An ant, walking on the plate, traverses a circle of radius 5 centered at the origin. What is the highest temperature encountered by the ant?

(A) 125
(B) 120
(C) 0
(D) 25
βœ“ Correct Answer: Option A Verified Solution

πŸ“˜ Step-by-Step Solution & Detailed Explanation:

The temperature function is given as: \[ T(x, y) = 4x^2 - 4xy + y^2 = (2x - y)^2 \] The goal is to maximize the function \( 2x - y \) while satisfying the constraint \( x^2 + y^2 = 25 \), which is the equation of a circle with radius 5 centered at the origin. Let \( 2x - y = k \). The maximum value of \( k \) will occur when the line \( 2x - y = k \) is tangent to the circle \( x^2 + y^2 = 25 \). The condition for tangency is derived using the distance between the center of the circle \((0, 0)\) and the line \( 2x - y = k \): \[ c^2 = a^2 (1 + m^2), \] where \( c \) is the distance from the center to the tangent line, \( a \) is the radius of the circle, and \( m \) is the slope of the line. Here, the slope \( m \) of the line \( 2x - y = k \) is \( m = 2 \). Substituting into the tangency condition: \[ k^2 = 25(1 + 2^2), \] \[ k^2 = 25(1 + 4) = 125. \] Thus, the maximum value of \( k \) is: \[ k = \pm \sqrt{125} = \pm 5\sqrt{5}. \] Therefore, the maximum (or minimum) value of \( 2x - y \) is \( \pm 5\sqrt{5} \).

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