The area of one triangle is 150 square centimeters when it's height is 20 centimeters and it's base length is 15 centimeters. Wh
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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
give us (x,y) values negatives, outside the region, so we do not take it in account
For : 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)