Answer to Question #114221 in Calculus for Pintoo

Question

Find the range of the function f defined by f(x y)=10-x^2-y^2 for all (x y) for which x^2+y^2<=9. Sketch two of its level curves.

Answer 1

Function $f$ is defined by $f(x, y)=10-x^2-y^2$ for all $(x, y)$ for which $x^2+y^2 \le 9.$ Also we know that for real numbers $x, y$ we get $0 \le x^2+y^2.$

Let us rewrite $f$ in the form $f(x ,y)=10-(x^2+y^2)$ .

Next we can write an inequation

$0\le x^2+y^2 \le 9.$

Therefore,

$10-0 \ge 10-(x^2+y^2) \ge 10-9.$

If we simplify the inequation above, we'll get

$10\ge 10-(x^2+y^2) \ge 1.$

So $1 \le f(x,y) \le 10.$

Let us discuss the properties of the range. We can see that points $(x,y)$ are situated inside the circle with center $(0,0)$ and radius $\sqrt{9}=3.$ This circle is shown by the black line in the figure below.

Now we'll consider the level curves. That means, we consider the set of points $(x,y)$ where $f(x,y)$ is a constant.

Let us consider the curve where $f(x,y) = 6$ . We should find points $(x,y)$ for which

$6=10-(x^2+y^2)$ , so $x^2+y^2=4.$ The curve is also a circle with center in (0,0) and radius $\sqrt{4}=2.$ This circle is shown by the blue line in the figure below.

Let us consider the curve where $f(x,y)=9$ . We should find points $(x,y)$ for which $9=10-(x^2+y^2)$, so $x^2+y^2=1.$ The curve is also a circle with center in (0,0) and radius $\sqrt{1}=1.$ This circle is shown by the red line in the figure below.