Mathematics, 18.07.2019 03:00 mememe87

Find the absolute maximum and minimum values of the function f(x, y) = x2 + xy + y2 on the disc x2 + y2 ≤ 1. (you do not have to use calculus.)

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Answer from: jojo10092003

The required absolute maximum and minimum values of the function is 0 and $\frac{3}{2}$ .

Given that,

Function $f(x, y) =x^{2} + xy+y^{2}$ on disc $x^{2} + y^{2} \leq 1$ .

We have to determine,

The absolute maximum and minimum values of the function.

According to the question,

Consider the polar points,

$x = rcos\theta, y = rsin\theta$

The equation of the circle, $x^{2} + y^{2} = 1$

Then, (x, y) can be written as,

$f(r,\theta) = r^{2} cos\theta sin\theta+r^{2} = \frac{r^{2} }{2} (sin2\theta+2)$

Therefore, $f(x, y) = f(r, \theta)$ which means they share the same extreme values.

Over the disk, $x^{2} + y^{2} \leq 1$ points lies $0\leq r\leq 1$

Where, r = 0 then $f(0, \theta) ]= 0$

And Where r = 1 then,

$f(1, \theta) = \frac{1}{2} sin2\theta+1$

When $-1\leq sin2\theta\leq 1$

Then,

$= -1\leq sin2\theta\leq 1\\\\= \frac{-1}{2}\leq \frac{1}{2}sin2\theta \leq \frac{1}{2} \\\\= \frac{-1}{2}+1\leq \frac{1}{2}sin2\theta+1 \leq \frac{1}{2}+1 \\\\= \frac{1}{2}\leq \frac{1}{2}sin2\theta \leq \frac{3}{2} \\$

Along the boundary $x^{2}+ y^{2} =1$ we get $\frac{1}{2} \leq f(r, \theta)\leq \frac{3}{2}$

Therefore, f(x, y) has a minimum value of 0 and a maximum value $\frac{3}{2}$ of over the unit disk $x^{2} + y^{2} \leq 1$ .

Hence, The required absolute maximum and minimum values of the function is 0 and $\frac{3}{2}$ .

For more information about the Maxima and Minima click the link given below.

link

Answer from: channarlawassociate

Consider converting to polar coordinates, setting

$x=r\cos\theta$
$y=r\sin\theta$
$\implies x^2+y^2=r^2$

Then f(x,y)

can be written as

$F(r,\theta)=r^2\cos\theta\sin\theta+r^2=\dfrac12r^2(\sin2\theta+2)$

i.e. $f(x,y)=F(r,\theta)$ , which means they share the same extreme values.

Over the disk $x^2+y^2\le1$ , it's easy to see that we would have $0\le r\le1$ .

In the least case, when r=0

, we have $F(0,\theta)=0$ .

At most, when r=1

, we have $F(1,\theta)=\dfrac12\sin2\theta+1$ . Since $-1\le\sin2\theta\le1$ , it follows that

$-\dfrac12\le\dfrac12\sin2\theta\le\dfrac12$
$\dfrac12\le\dfrac12\sin2\theta+1\le\dfrac32$

so that, along the boundary x^2+y^2=1

, we get $\dfrac12\le F(r,\theta)\le\dfrac32$ .

So we found that f(x,y)

has a minimum value of 0 and a maximum value of $\dfrac32$ over the unit disk $x^2+y^2\le1$ .

Answer from: Quest

the degrees would remain the same ( the number to the left of the decimal point.

the multiply the decimal by 60 to get the minutes and use the numbers to the left of the decimal point.

then multiply the decimal portion of that by 60 for the seconds and round off as needed.

39 .956652

the 39 remains the degrees.

the minutes = 0.956652 x 60 = 57.39912

the minutes = 57

the seconds = 0.39912 x 60 = 23.95 = 24 ( rounded to nearest tenth)

so 39.956652 = 39 degrees, 57 minutes and 24 seconds

-75.159809

the -75 remains the degrees

the minutes = 0.159809 x 60 = 9.58854

the minutes = 9

the seconds = 0.58854 x 60 = 35.31, rounded to nearest tenth = 35.3 seconds

so -75.159809 = -75 degrees, 9 minutes and 35.3 seconds.

the first coordinate is north or south, since it it positive it is north.

the second number is east or west, because it is negative it is west.

39 degrees, 57 minutes and 24 seconds n

-75 degrees, 9 minutes and 35.3 seconds w

Answer from: Quest

step by step resolutions

1) n+n+1+n+2=51

3n+3=51

3n=48

n=16

the three numbers are 16, 17 and 18.

2) x+15=4x

3x=15

x=5

radha is now 5 years old

3) the two digits are x and y.

y=3x

(10x+y)+(10y+x)=88

y=3x

11x+11y=88

y=3x

x+y=8

x+3x=8

x=2

the original number can be either 28 or 82.

4) n+(n+2)+(n+4)=48

3n+6=48

3n=42

n=14

the numbers are 14, 16 and 18.

5) y=x+3

10y+x+10x+y=143

y=x+3

11x+11y=143

y=x+3

x+y=13

x+x+3=13

2x=10

x=5

the original number can be 57 or 75.

hope this !

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Find the absolute maximum and minimum values of the function f(x, y) = x2 + xy + y2 on the disc x2 +...

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