Question

Calc III homework probs (with a friend) Calc III homework probs (with a friend) @Mathematics

Answer 1

f(x,y)=x^3(y)-xy+y^3(x) find all crit points and classify each as a relative max, min, or saddle point using the second partials test.

Answer 2

find the first partials first right? So you have:
\[f_x(x,y)=3yx^2-y+y^3; f_y(x,y)=x^3-x+3xy^2\]
Set those to zero.
\[0=y(3x^2-1+y^2); x(x^2-1+3y^2)=0\]
So if x=0, then y=0 holds.
Then we have:\[3x^2-1+y^2=0 \implies x^2=\frac{1}{3}(1-y^2) \implies x=\pm \sqrt{\frac{1}{3}(1-y^2)}\]
From this we can see that if that holds then:
\[\frac{1}{3}(1-y^2)-1+3y^2=0 \implies -\frac{2}{3}+\frac{5}{3}y^2=0 \implies 5y^2=2 \implies y=\pm \sqrt{\frac{2}{5}}\]
Which means that:
\[x=\pm \sqrt{\frac{1}{5}}\]
Then calculate the derivatives and plug those in. Done.

Answer 3

\[y=0; x=\pm 1 \]
Also works.

Answer 4

x=0, y=pm 1 works

Answer 5

Find all critical points of the function and determine the absolute max and mins on the region R.

\[\frac{24xy}{(x^2+4)(y^2+9)}\]

Answer 6

R = {(x,y):0<=x<=1 and 0<=y<=5}

Answer 7

Just find the critical points like normal and then make sure they fall in the range r
its not hard.