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 October 21st, 2013, 06:21 AM #1 Newbie   Joined: Aug 2013 Posts: 1 Thanks: 0 Find the absolute minimum and absolute maximum. Find the absolute minimum and absolute maximum of f(x,y) = x^2+y^2-2(x-y) in the region defined by the x^2+y^2?4 ?
 October 21st, 2013, 07:03 AM #2 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Find the absolute minimum and absolute maximum. When you're looking for minima and maxima, the first thing to try is to differentiate. In this case, as f is afunction of x and y, you need to take partial derivatives wrt x and y. $\frac{\partial f}{\partial x}= 2x - 2, \ and \ \frac{\partial f}{\partial y} = 2y + 2$ So, the partial derivatives of f are 0 when x = 1 and y = -1. And, as both second derivatives are positive, these represent local minima, hence an absolute minimum for f at (1, -1). The maxima must then occur on the boundary, as there are no local maxima for f. On the boundary, we have: $f(x,y)= 4 - 2x + 2y, \ where \ x^2 + y^2 = 4$ You can you can use more calculus to show that a maximum occurs at: $x= -\sqrt{2}, \ y = \sqrt{2}$
October 21st, 2013, 06:53 PM   #3
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Re: Find the absolute minimum and absolute maximum.

Quote:
 Originally Posted by Pero When you're looking for minima and maxima, the first thing to try is to differentiate. In this case, as f is afunction of x and y, you need to take partial derivatives wrt x and y. $\frac{\partial f}{\partial x}= 2x - 2, \ and \ \frac{\partial f}{\partial y} = 2y + 2$ So, the partial derivatives of f are 0 when x = 1 and y = -1. And, as both second derivatives are positive, these represent local minima, hence an absolute minimum for f at (1, -1). The maxima must then occur on the boundary, as there are no local maxima for f. On the boundary, we have: $f(x,y)= 4 - 2x + 2y, \ where \ x^2 + y^2 = 4$
I had to think about that for a second! Pero replaced the first "$x^2+ y^2$" in $x^2+ y^2- 2(x- y)$ with 4. Now there are several ways to go to "use more calculus" as Pero says below:
We could replace y with $\pm\sqrt{4- x^2}$. That would be the hard way!

We could switch to polar coordinates with r= 2 (since $x^2+ y^2= 4$ is a circle with center at (0, 0) and radius 2) to have $x= 2cos(\theta)$ and $y= 2sin(\theta)$ so we are finding min and max of $4- 4cos(\theta)+ 4 sin(\theta)$ for $0\le \theta\le 2\pi$. That's fairly straight forward.

We could use "Lagrange Multipliers" to minimize f(x,y)= 4- 2x+ 2y subject to the condition that $g(x, y)= x^2+ y^2= 4$. $\nabla f= -2\vec{i}+ 2\vec{j}$ and $\nabla g= 2x\vec{i}+ 2y\vec{j}$. We must have $\nabla f= \lambda\nabla g$ for some number $\nabla$ so we must have $-2= 2\lambda x$ and $2= 2\lambda y$, together with $x^2+ y^2= 4$, three equations to solve for the three unknowns, x, y, and $\lambda$. Since a specific value of $\lambda$ is not necessary for a solution, I find it is often best to eliminate $\lambda$ first by dividing one equation by another. Here, that gives $\frac{-2}{2}= \frac{2\lambda x}{2\lambda y}$ which reduces to $-1= \frac{x}{y}$ so y= -x. Together with $x^2+ y^2= 4$, we have $x^2+ (-x^2)^2= 2x^2= 4$ so that $x= \pm\sqrt{2}$. Together with y= -x, we have two solutions, $\left(\sqrt{2}, -\sqrt{2}\right)$ and $\left(-\sqrt{2}, \sqrt{2}\right)$.

 October 22nd, 2013, 12:38 AM #4 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Find the absolute minimum and absolute maximum. It might be useful to remember that on a circle x+y has a maximum when $x=y=\sqrt{2}$. In an exam, you'll have to work this out every time(and using polar coordinates is the best way to go), but it's the sort of thing I just remember. In this case we had to maximise y-x, so that would be at $(-\sqrt{2}, \sqrt{2})$ While $(\sqrt{2}, -\sqrt{2})$ is the minimum on the boundary. Also: $sin(x) + cos(x) \ has \ a \ maximum \ at \ x=\frac{\pi}{4}$
 October 22nd, 2013, 01:29 AM #5 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 Re: Find the absolute minimum and absolute maximum. When they say 'absolute minimum' are they referring to 'global minimum' or 'minimizer of the absolute value'? I believe they're referring to the second.
 October 22nd, 2013, 01:31 AM #6 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 Re: Find the absolute minimum and absolute maximum. My bad, they're referring to global extrema, which is the first. I just checked.

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