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

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

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).

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

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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Grigori, You have an error when differentiating with respect to x:

f'=4-3x/sqrt(1-x2)

Using this and following through with your solution you will get max value of f(x,y) on the circle is 5.

You are right. This was an accident. I was doing part of my calculations mentally and forgot to cancel two 2-s, in the numerator and denominator. If you solve my equations after these corrections you will come with x = 4/5 and y = 3/5

and the maximum will be  f(x) = 5 as found by Robert. He did very good job and was more accurate with numerical calculations.

You are right. This was an accident. I was doing part of my calculations mentally and forgot to cancel two 2-s, in the numerator and denominator. If you solve my equations after these corrections you will come with x = 4/5 and y = 3/5

and the maximum will be  f(x) = 5 as found by Robert. He did very good job and was more accurate with numerical calculations.

You are right. This was an accident. I was doing part of my calculations mentally and forgot to cancel two 2-s, in the numerator and denominator. If you solve my equations after these corrections you will come with x = 4/5 and y = 3/5

and the maximum will be  f(x) = 5 as found by Robert. He did very good job and was more accurate with numerical calculations.