Find the maximal value of f(x, y)?

Question

Find the maximal value of f(x, y)=3y+4x on the circle x^2+y^2=1.

Answer 1

Method I. Using linear programming concept

Since the maximum value must be reached at the boundary of the circle,  draw a tangent line with the circle such that the tangent line has a slope of -4/3. Therefore, the point of tangency can be written as (4a, 3a), where a > 0. Plug into the equation of the circle: (3a)^2 + (4a)^2 = 1 => a = 1/5

fmax = f(4/5, 3/5) = 3(3/5) + 4(4/5) = 5 <==Answer

 

Method II. Using Lagrange multiplier

g(x, y) = 3y+4x + λ(x^2+y^2-1)

g'_x = 4 + 2λx = 0 => x = -2/λ, (x > 0)

g'_y = 3 + 2λy = 0 => y = -3/(2λ), (y > 0)

x^2+y^2 = (-2/λ)^2 + (-3/(2λ))^2 = 1 => λ = -2.5

x = 4/5, y = 3/5

fmax = f(4/5, 3/5) = 3(3/5) + 4(4/5) = 5 <==Answer

Answer 2

Find y from the equation for the circle: y = sqrt(1-x²⁾ and rewrite the function f(x,y) in terms of x only:

  f(x,(y(x)) = 3sqrt(1-x²) + 4x. Differentiate by x. It gives us

                                                       df(x)/dx = 4 - 6x/sqrt(1-x²)

if f(x) has a maximum on a circle, then df/dx = 0 and the derivative changes its sign from positive to negative after passing that point x. Puting df/dx = 0 we obtain

                                                               4 = 6/sqrt(1-x²)

raising both sides into square we will find    x = 4/sqrt(52)   and y = 6/sqrt(52). For x < 4/sqrt(52) the derivarive is negative, and for x>4/sqrt(52) is positive. The naximum f(xx,y) is

                       3x6/sqrt(52) + 4x4/sqrt(52) = 34/sqrt(52) = 4.71

          

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