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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.

Comments

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).

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2 Answers

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

          

                 .

Comments

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.

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