Minus always used ____

Find the absolute maximum and absolute minimum values of the function f(x, y) = x 2 + y 2 − x 2 y + 7 on the set d = {(x, y) : |

dsp73

Looks like $f(x,y)=x^2+y^2-x^2y+7$ .

$f_x=2x-2xy=0\implies2x(1-y)=0\implies x=0\text{ or }y=1$

$f_y=2y-x^2=0\implies2y=x^2$

If $x=0$ , then $y=0$ - critical point at (0, 0).
If $y=1$ , then $x=\pm\sqrt2$ - two critical points at $(-\sqrt2,1)$ and $(\sqrt2,1)$

The latter two critical points occur outside of $D$ since $|\pm\sqrt2|>1$ so we ignore those points.

The Hessian matrix for this function is

$H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}2-2y&-2x\\-2x&2\end{bmatrix}$

The value of its determinant at (0, 0) is $\det H(0,0)=4>0$ , which means a minimum occurs at the point, and we have $f(0,0)=7$ .

Now consider each boundary:

If $x=1$ , then

$f(1,y)=8-y+y^2=\left(y-\dfrac12\right)^2+\dfrac{31}4$

which has 3 extreme values over the interval $-1\le y\le1$ of 31/4 = 7.75 at the point (1, 1/2); 8 at (1, 1); and 10 at (1, -1).

If $x=-1$ , then

$f(-1,y)=8-y+y^2$

and we get the same extrema as in the previous case: 8 at (-1, 1), and 10 at (-1, -1).

If $y=1$ , then

$f(x,1)=8$

which doesn't tell us about anything we don't already know (namely that 8 is an extreme value).

If $y=-1$ , then

$f(x,-1)=2x^2+8$

which has 3 extreme values, but the previous cases already include them.

Hence $f(x,y)$ has absolute maxima of 10 at the points (1, -1) and (-1, -1) and an absolute minimum of 0 at (0, 0).

3 0

3 years ago

Find the perimeter and the area of the polygon with the given vertices.

7nadin3 [17]

Answer:
Area : 45 Perimeter : 2,025

Explanation:
Area: 5 times 9 equals 45. The 5 is the polygon’s base, the 9 is the polygon’s height.
Perimeter: 5 times 5 times 9 times 9 equals 2,025. Perimeter is base times base times height times height.

Glad I could help! :)

3 0

3 years ago

Simplify the expression (8x^3y^2/4a^2b^4)^-2

icang [17]

Simplify the following:
((8 x^3 y^2 a^2 b^4)/4)^(-2)
8/4 = (4×2)/4 = 2:
(2 x^3 y^2 a^2 b^4)^(-2)
Multiply each exponent in 2 x^3 y^2 a^2 b^4 by -2:
(x^(-2×3) y^(-2×2) a^(-2×2) b^(-2×4))/(2^2)
-2×3 = -6:
(x^(-6) y^(-2×2) a^(-2×2) b^(-2×4))/2^2
-2×2 = -4:
(y^(-4) a^(-2×2) b^(-2×4))/(2^2 x^6)
-2×2 = -4:
(a^(-4) b^(-2×4))/(2^2 x^6 y^4)
-2×4 = -8:
b^(-8)/(2^2 x^6 y^4 a^4)
2^2 = 4:
Answer: 1/(4 x^6 y^4 a^4 b^8) thus the answer is A

3 0

3 years ago

What is 347500 rounded to the nearest 1000

Annette [7]

It is 348000 because if its 5 or above it rounds up and 4 or below rounds down so the 5 rounds the 7 to an 8

5 0

3 years ago

Sari sold 300 boxes of greeting cards in November. In december she sold 500. What was the percentage increase in her sales

grigory [225]

There was a 66 $\frac{2}{3}$ % increase from Sari's sales from November to December.

To find the percentage increase in her sales, we must first find the amount her sales increased. We can do that by subtracting the greeting cards sold in November from the number of greeting cards sold in December. Substituting values and simplifying, we get that she sold 200 more greeting cards in December.

To find what percent increase this is, we must find what percent 200 is of the number of greeting cards sold in November. We can do this by dividing 200 by the number of cards sold in November (300), and then multiplying by 100 (to make it a percent). Doing this, we get $\frac{2}{3}$ * 100 which gives us our answer of 66 $\frac{2}{3}$ %.

3 0

3 years ago

Minus always used _____ a term