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Find the conditional extrema of $$f(x,y)=\cos^2x+\cos^2y,\quad g(x,y)=x-y+\frac{\pi}{4}=0.$$

I have a problem with finding a solution to this problem. Using Lagrange multipliers i come up with a system of three equations which i don't know how to solve: \begin{align} F_x &=-2\sin x\cos x+1=0\\ F_y &=-2\sin y\cos y-1=0\\ F_\lambda &=x-y+\frac{\pi}{4}=0 \end{align}

Any help will be appreciated.

Leucippus
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Mattematics
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  • The first two equations are missing a $\lambda$. After the fix, you will be looking at $-\sin 2x=-\lambda$, $-\sin 2y=\lambda$. After dealing with the special case $\lambda=0$, you get $\sin 2x=-\sin 2y$. So $2x$ and $2y$ are close relatives. – André Nicolas Aug 04 '15 at 15:15

2 Answers2

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Your $F_x$ and $F_y$ are wrong. Since $$F(x,y,\lambda) = \cos^2 x + \cos^2 y +\lambda (x-y+\frac\pi4)$$

Derivatives should be $$F_x=-2\cos x \sin x + \lambda\\ F_y = -2\cos y \sin y - \lambda$$

Then, I suggest you sum the first two equations ($F_x=0$ and $F_y=0$).

5xum
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HINT:

W/O using calculus,

Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,

$$\cos^2x+\cos^2y=\cdots=1+\cos(x-y)\cos(x+y)=1+\cos\left(-\dfrac\pi4\right)\cos\left(y-\dfrac\pi4+y\right)$$

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lab bhattacharjee
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