Respuesta :
Consider converting to polar coordinates, setting
[tex]x=r\cos\theta[/tex]
[tex]y=r\sin\theta[/tex]
[tex]\implies x^2+y^2=r^2[/tex]
Then [tex]f(x,y)[/tex] can be written as
[tex]F(r,\theta)=r^2\cos\theta\sin\theta+r^2=\dfrac12r^2(\sin2\theta+2)[/tex]
i.e. [tex]f(x,y)=F(r,\theta)[/tex], which means they share the same extreme values.
Over the disk [tex]x^2+y^2\le1[/tex], it's easy to see that we would have [tex]0\le r\le1[/tex].
In the least case, when [tex]r=0[/tex], we have [tex]F(0,\theta)=0[/tex].
At most, when [tex]r=1[/tex], we have [tex]F(1,\theta)=\dfrac12\sin2\theta+1[/tex]. Since [tex]-1\le\sin2\theta\le1[/tex], it follows that
[tex]-\dfrac12\le\dfrac12\sin2\theta\le\dfrac12[/tex]
[tex]\dfrac12\le\dfrac12\sin2\theta+1\le\dfrac32[/tex]
so that, along the boundary [tex]x^2+y^2=1[/tex], we get [tex]\dfrac12\le F(r,\theta)\le\dfrac32[/tex].
So we found that [tex]f(x,y)[/tex] has a minimum value of 0 and a maximum value of [tex]\dfrac32[/tex] over the unit disk [tex]x^2+y^2\le1[/tex].
[tex]x=r\cos\theta[/tex]
[tex]y=r\sin\theta[/tex]
[tex]\implies x^2+y^2=r^2[/tex]
Then [tex]f(x,y)[/tex] can be written as
[tex]F(r,\theta)=r^2\cos\theta\sin\theta+r^2=\dfrac12r^2(\sin2\theta+2)[/tex]
i.e. [tex]f(x,y)=F(r,\theta)[/tex], which means they share the same extreme values.
Over the disk [tex]x^2+y^2\le1[/tex], it's easy to see that we would have [tex]0\le r\le1[/tex].
In the least case, when [tex]r=0[/tex], we have [tex]F(0,\theta)=0[/tex].
At most, when [tex]r=1[/tex], we have [tex]F(1,\theta)=\dfrac12\sin2\theta+1[/tex]. Since [tex]-1\le\sin2\theta\le1[/tex], it follows that
[tex]-\dfrac12\le\dfrac12\sin2\theta\le\dfrac12[/tex]
[tex]\dfrac12\le\dfrac12\sin2\theta+1\le\dfrac32[/tex]
so that, along the boundary [tex]x^2+y^2=1[/tex], we get [tex]\dfrac12\le F(r,\theta)\le\dfrac32[/tex].
So we found that [tex]f(x,y)[/tex] has a minimum value of 0 and a maximum value of [tex]\dfrac32[/tex] over the unit disk [tex]x^2+y^2\le1[/tex].
The required absolute maximum and minimum values of the function is 0 and [tex]\frac{3}{2}[/tex].
Given that,
Function [tex]f(x, y) =x^{2} + xy+y^{2}[/tex] on disc [tex]x^{2} + y^{2} \leq 1[/tex] .
We have to determine,
The absolute maximum and minimum values of the function.
According to the question,
Consider the polar points,
[tex]x = rcos\theta, y = rsin\theta[/tex]
The equation of the circle, [tex]x^{2} + y^{2} = 1[/tex]
Then, (x, y) can be written as,
[tex]f(r,\theta) = r^{2} cos\theta sin\theta+r^{2} = \frac{r^{2} }{2} (sin2\theta+2)[/tex]
Therefore, [tex]f(x, y) = f(r, \theta)[/tex] which means they share the same extreme values.
Over the disk, [tex]x^{2} + y^{2} \leq 1[/tex] points lies [tex]0\leq r\leq 1[/tex]
Where, r = 0 then [tex]f(0, \theta) ]= 0[/tex]
And Where r = 1 then,
[tex]f(1, \theta) = \frac{1}{2} sin2\theta+1[/tex]
When [tex]-1\leq sin2\theta\leq 1[/tex]
Then,
[tex]= -1\leq sin2\theta\leq 1\\\\= \frac{-1}{2}\leq \frac{1}{2}sin2\theta \leq \frac{1}{2} \\\\= \frac{-1}{2}+1\leq \frac{1}{2}sin2\theta+1 \leq \frac{1}{2}+1 \\\\= \frac{1}{2}\leq \frac{1}{2}sin2\theta \leq \frac{3}{2} \\[/tex]
Along the boundary [tex]x^{2}+ y^{2} =1[/tex] we get [tex]\frac{1}{2} \leq f(r, \theta)\leq \frac{3}{2}[/tex]
Therefore, f(x, y) has a minimum value of 0 and a maximum value [tex]\frac{3}{2}[/tex] of over the unit disk [tex]x^{2} + y^{2} \leq 1[/tex] .
Hence, The required absolute maximum and minimum values of the function is 0 and [tex]\frac{3}{2}[/tex].
For more information about the Maxima and Minima click the link given below.
https://brainly.com/question/12982238
