Question

Here is a contour plot of the function f(x,y) = 4 + x + y - 3xy. (Click the image to enlarge it.) By looking at the contour plot, characterize the two critical points of the function. You should be able to do this analysis without computing derivatives, but you may want to compute them to corroborate your intuition The critical point (1.1) is a ??? (choose one from the list). The second critical point is at the point and it is a ???

Answer 1

The critical point (1,1) is a local/relative minimum

The second critical point is at the point (0,0) and it is a saddle point

Explanation:

`\displaystyle f(x,y)=4+x^3+y^3-3xy\\\\(\partial f)/(\partial x)=3x^2-3y\\\\(\partial f)/(\partial y)=3y^2-3x`

`3x^2-3y=0\\3x^2=3y\\x^2=y\\\\3y^2-3x=0\\3(x^2)^2-3x=0\\3x^4-3x=0\\x^4-x=0\\x(x^3-1)=0\\x(x-1)(x^2+x+1)=0\\\\x=0,\,1 \text{ only real critical points}`

When x=0

`\displaystyle H=\biggr((\partial^2 f)/(\partial x^2)\biggr)\biggr((\partial^2 f)/(\partial y^2)\biggr)-\biggr((\partial^2 f)/(\partial x\partial y)\biggr)^2\\\\H=(6x)(6y)-(-3)^2\\\\H=(6x)(6y)-9\\\\H=(6\cdot0)(6\cdot 0)-9\\\\H=-9 < 0`

Therefore, (0,0) is a saddle point

When x=1

`H=(6\cdot1)(6\cdot 1)-9\\H=36-9\\H=27 > 0`

Since
`(\partial^2 f)/(\partial x^2) > 0`, then (1,1) is a local minimum