Can someone please help me with this

Find the absolute maximum and minimum values of f(x, y) = x+y+ p 1 − x 2 − y 2 on the quarter disc {(x, y) | x ≥ 0, y ≥ 0, x2 +

Andreas93 [3]

Answer:

absolute max: f(x,y)=1/2+p1 ; at P(1/2,1/2)

absolute min: f(x,y)=p1 ; at U(0,0), V(1,0) and W(0,1)

Step-by-step explanation:

In order to find the absolute max and min, we need to analyse the region inside the quarter disc and the region at the limit of the disc:

Region inside the quarter disc:

There could be Minimums and Maximums, if:

∇f(x,y)=(0,0) (gradient)

we develop:

(1-2x, 1-2y)=(0,0)

x=1/2

y=1/2

Critic point P(1/2,1/2) is inside the quarter disc.

f(P)=1/2+1/2+p1-1/4-1/4=1/2+p1

f(0,0)=p1

We see that:

f(P)>f(0,0), then P(1/2,1/2) is a maximum relative

Region at the limit of the disc:

We use the Method of Lagrange Multipliers, when we need to find a max o min from a f(x,y) subject to a constraint g(x,y); g(x,y)=K (constant). In our case the constraint are the curves of the quarter disc:

g1(x, y)=x^2+y^2=1

g2(x, y)=x=0

g3(x, y)=y=0

We can obtain the critical points (maximums and minimums) subject to the constraint by solving the system of equations:

∇f(x,y)=λ∇g(x,y) ; (gradient)

g(x,y)=K

Analyse in g2:

x=0;

1-2y=0;

y=1/2

Q(0,1/2) critical point

f(Q)=1/4+p1

We do the same reflexion as for P. Q is a maximum relative

Analyse in g3:

y=0;

1-2x=0;

x=1/2

R(1/2,0) critical point

f(R)=1/4+p1

We do the same reflexion as for P. R is a maximum relative

Analyse in g1:

(1-2x, 1-2y)=λ(2x,2y)

x^2+y^2=1

Developing:

x=1/(2λ+2)

y=1/(2λ+2)

x^2+y^2=1

So:

(1/(2λ+2))^2+(1/(2λ+2))^2=1

$\lambda_{1}=\sqrt{1/2}*-1 =-0.29$

$\lambda_{2}=-\sqrt{1/2}*-1 =-1.71$

$\lambda_{2}$ give us (x,y) values negatives, outside the region, so we do not take it in account

For $\lambda_{1}$ : S(x,y)=(0.70, 070)

and

f(S)=0.70+0.70+p1-0.70^2-0.70^2=0.42+p1

We do the same reflexion as for P. S is a maximum relative

Points limits between g1, g2 y g3

we need also to analyse the points limits between g1, g2 y g3, that means U(0,0), V(1,0), W(0,1)

f(U)=p1

f(V)=p1

f(W)=p1

We can see that this 3 points are minimums relatives.

Conclusion:

We compare all the critical points P,Q,R,S,T,U,V,W an their respective values f(x,y). We find that:

absolute max: f(x,y)=1/2+p1 ; at P(1/2,1/2)

absolute min: f(x,y)=p1 ; at U(0,0), V(1,0) and W(0,1)

4 0

3 years ago

PLS HELP THIS IS HARD ANYONE PLS

aalyn [17]

Answer:

a or c

Step-by-step explanation:

4 0

3 years ago

Write in standard form an equation for the line with slope 1/3 through the point (2,-2).

Alexxandr [17]

Answer:

x - 3y = 8.

Step-by-step explanation:

Use the point-slope form of the equation of a line:

y - y1 = m(x - x1) where m = the slope and (x1, y1) is a point on the line.

So substituting the given values:

y - (-2) = 1/3(x - 2)

y + 2 = 1/3x - 2/3 Multiply through by 3:

3y + 6 = x - 2

x - 3y = 6 + 2

x - 3y = 8 <---- Standard Form.

6 0

3 years ago

Find the sum of 5m + 3n + p, -5p + 3n, and 2n - m

tigry1 [53]

5m + 3n + p -5p + 3n + 2n - m
= 4m + 8n - 4p (combine like terms)

hope it helps

7 0

3 years ago

Read 2 more answers

What do you think Ginsburg is saying about a judge's role in the Supreme Court?

VLD [36.1K]

$\huge\mathfrak\red{Answer}$

Ruth Bader Ginsburg began her career as a justice where she left off as an advocate, fighting for women's rights. In 1996, Ginsburg wrote the majority opinion in United States v. Virginia, holding that qualified women could not be denied admission to Virginia Military Institute.

✔❤

4 0

3 years ago