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

## Comments

But how did you find the slope, which is -4/3? And how did you find (4a, 3a)?

3y+4x = c

The slope of the line is -4/3

At the point of tangency on the circle, the slope must be 3/4 such that the radius is perpendicular to the tangent line. So, we can assume the point at (4a, 3a).