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BartSMP [9]
2 years ago
11

Someone help me p,ease as soon as possible

Mathematics
1 answer:
just olya [345]2 years ago
8 0

1st hen =6x

2nd hen =7y

 1st hen is greater than 2nd

 so you get 6x > 7y

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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:

<u>Region inside the quarter disc:</u>

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

<u>Region at the limit of the disc:</u>

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

<u>Analyse in g2:</u>

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

<u>Analyse in g3:</u>

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

<u>Analyse in g1:</u>

(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

<u>Points limits between g1, g2 y g3</u>

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.

<u>Conclusion:</u>

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
Will select brainliest if ur answer is correct
ella [17]

Answer:

B (I think)

Step-by-step explanation:

A function can't have two X values that are the same (If I remember correctly) and the second table is the only one that doesn't have duplicated X values

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2 years ago
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scoray [572]
All you do is plug in -7 so you get f(x)=8(-7)*15 you get -56*15 which goes to -840
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Kisachek [45]
The answer is (4p+3)^2
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Determine the measure of ∠Q, rounded to the nearest tenth of a degree.<br> images attached.
const2013 [10]

Answer:

D

Step-by-step explanation:

Using the Sine rule in Δ QRS , that is

\frac{q}{sinQ} = \frac{r}{sinR} , that is

\frac{15}{sinQ} = \frac{12}{sin50.1} ( cross- multiply )

12× sinQ = 15 × sin50.1° ( divide both sides by 12 )

sinQ = \frac{15sin50.1}{12} , thus

Q = sin^{-1} ( \frac{15sin50.1}{12} ) ≈ 73.5° ( to the nearest tenth )

8 0
3 years ago
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