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#### Petrus

##### Well-known member

- Feb 21, 2013

- 739

Calculate highest and lowest value of the function

\(\displaystyle f(x,y)\frac{6}{x^2+y^2+1}+3xy\) in the range

\(\displaystyle \frac{1}{3}≤x^2+y^2≤2\)

Progress

\(\displaystyle f_x(x,y)=3y-\frac{12x}{(x^2+y^2+1)^2}\)

\(\displaystyle f_y(x,y)=3x-\frac{12y}{(x^2+y^2+1)^2}\)

Now when I have derivate respect to x or y I shall find critical point.

\(\displaystyle 3y-\frac{12x}{(x^2+y^2+1)^2}=0\)

\(\displaystyle 3x-\frac{12y}{(x^2+y^2+1)^2}=0\)

Well for the numerator:

\(\displaystyle 3x-12y=0<=>x=4y\)

\(\displaystyle 3y-12x=0<=>3y-48y=0 <=>y=0\) and that means also \(\displaystyle x=0\)

Denominator:

\(\displaystyle 1-(x^2+y^2+1)^2=0 <=>x=\sqrt{+-1-y^2-1^2}\) (I will admit it did take me a while notice I had 1 minus the function )

well to input that x into the function is the part I am unsure too handle with.

\(\displaystyle f(x,y)\frac{6}{x^2+y^2+1}+3xy\) in the range

\(\displaystyle \frac{1}{3}≤x^2+y^2≤2\)

Progress

\(\displaystyle f_x(x,y)=3y-\frac{12x}{(x^2+y^2+1)^2}\)

\(\displaystyle f_y(x,y)=3x-\frac{12y}{(x^2+y^2+1)^2}\)

Now when I have derivate respect to x or y I shall find critical point.

\(\displaystyle 3y-\frac{12x}{(x^2+y^2+1)^2}=0\)

\(\displaystyle 3x-\frac{12y}{(x^2+y^2+1)^2}=0\)

Well for the numerator:

\(\displaystyle 3x-12y=0<=>x=4y\)

\(\displaystyle 3y-12x=0<=>3y-48y=0 <=>y=0\) and that means also \(\displaystyle x=0\)

Denominator:

\(\displaystyle 1-(x^2+y^2+1)^2=0 <=>x=\sqrt{+-1-y^2-1^2}\) (I will admit it did take me a while notice I had 1 minus the function )

well to input that x into the function is the part I am unsure too handle with.

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