Lagrange Multipliers to find the Maximum and Minimum values

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The discussion focuses on using Lagrange Multipliers to find the maximum and minimum values of the function f(x,y) = x² - y, subject to the constraint g(x,y) = x² + y² = 25. The gradients of both functions were calculated correctly, leading to the simultaneous equations needed for solving the problem. There was initial confusion regarding the formulation of the equations, specifically whether to use 2x = λ2y, but clarification confirmed that the correct approach was followed. The final solutions were derived, identifying the maximum value at f(sqrt(25-y²), -1/2) = 101/4 and the minimum at f(-sqrt(25-y²), -1/2) = -97/4. The discussion concludes with a correction of a sign error in the calculation of x, ensuring accurate results.
brendan
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Homework Statement



Use Lagrange Multipliers to find the Maximum and Minimum values of f(x,y) = x2-y.
Subject to the restraint g(x,y) = x2+y2=25

Homework Equations



gradient f(x,y)= gradient g(x,y)

The Attempt at a Solution



I have found the gradients of f and g to be

f(x,y) = 2xi + -1j
g(x,y) = 2xi + 2yj


I have put these to gether with the constraint to find the simutaneous equation

2x = lambda 2x
-1 = lambda 2y
x2+y2 = 25

Have I got this system right so far ?

I'm a bit concerned about the 2x = lambda 2x part.
I'm not sure if this is right or I should have used 2x = lambda 2y ?

In the original system I have Lambda = 1 and y = -1/2

Any help greatly appreciated
Brendan
 
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brendan said:
I'm a bit concerned about the 2x = lambda 2x part.
I'm not sure if this is right or I should have used 2x = lambda 2y?

Everything looks right to me, and no, you definitely should not have used 2x=\lambda 2y. The general system for a function f(x,y) constrained to g(x,y) = k is:
\begin{align*}<br /> f_x &amp;= \lambda g_x \\<br /> f_y &amp;= \lambda g_y \\<br /> g &amp;= k<br /> \end{align*}
So you have done this exactly right. You have found \lambda=1,y=-1/2 correctly. What about x?
 
Last edited:
What are you concerned about? I think you are doing fine. 2x = lambda 2y isn't right at all. Why would you think it is?
 
g(x,y) = x2+y2=25

Now we have y = -1/2

so x = sqrt(y2+25) or -sqrt(y2+25)

= sqrt(101)/2 or -sqrt(101)


So we have two solutions
f(x,y)= x2-y = f(sqrt(101)/2,-1/2) = 103/4 = 25.75
f(x,y)= x2-y = f(-sqrt(101)/2,-1/2) = 103/4 = 25.75

So they are both the maximum
 
brendan said:
g(x,y) = x2+y2=25

Now we have y = -1/2

so x = sqrt(y2+25) or -sqrt(y2+25)

Whoops, you have a sign error. It should have been x = \pm\sqrt{25-y^2}.
 
g(x,y) = x2+y2=25

Now we have y = -1/2

so x = sqrt(25-y2) or -sqrt(25-y2)

= sqrt(99/4) or -sqrt(99/4)


So we have two solutions
f(x,y)= x2-y = fsqrt(25-y2) ,-1/2) = 101/4= 25.25 (maximum)
f(x,y)= x2-y = f(-sqrt(25-y2) ,-1/2) = -97/4= -24.25 (Minimum)

Is that better?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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