Wikipedia:Reference desk/Archives/Mathematics/2011 February 24

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February 24

Partial differentiation quiz

Hello all! Today I had my partial differentiation quiz. There was a problem that I didn't get. It went something like: "Using the partial and total derivatives, find the range (there might have been another term than range, the meaning was the range of z values) of the function z(x,y)=sin(x)+cos(y)-sin(x)cos(y)". I knew that the answer was {z: -3≤z≤1} from lower math reasoning (since sin and cos are bounded on R -1 to 1 and z is continuous) but I didn't know how to solve it using the method it was testing. I knew it had to be a maximization/minimization problem, but I got stuck on how to find the global maximums and minimums, especially since there are an infinite number of local minimums. How should I have solved this? Thanks. 72.128.95.0 (talk) 23:26, 24 February 2011 (UTC)

z is periodic in each direction with period $2\pi$, of course, so it suffices to consider (say) $x,y\in [0,2\pi )$. To get the extrema, you look for places where the gradient is 0; the gradient is $\langle (1-\cos y)\cos x,(\sin x-1)\sin y\rangle$. Both elements have to be 0 simultaneously, so you have $(\cos x=0\lor \cos y=1)\land (\sin y=0\lor \sin x=1)$. Applying the restriction on the ranges, that yields $\left(x={\frac {\pi }{2}}\lor x={\frac {3\pi }{2}}\lor y=0\right)\land \left(y=0\lor y=\pi \lor x={\frac {\pi }{2}}\right)$ which simplifies to $y=0\lor x={\frac {\pi }{2}}\lor (x,y)=\left({\frac {3\pi }{2}},\pi \right)$. Those in turn yield $z=1$, $z=1$, and $z=-3$. Since z is bounded, we know that the range must be the interval between them. --Tardis (talk) 00:05, 25 February 2011 (UTC)

There is also a simpler way to do it. Make the change of variables a = sin(x), b = cos(y). Then the new problem is to find the extrema of

$f(a,b)=a+b-ab,\quad (a,b)\in [-1,1]\times [-1,1].$

There are no critical points in the interior of the square, so it's enough to solve the optimization problem on the boundary. We can pretty much do this by inspection, giving $f(-1,-1)=-3$ for the minimum and $f(0,1)=1$ for the maximum. Sławomir Biały (talk) 00:17, 25 February 2011 (UTC)