Respuesta :
Answer:
the maximum values of f is f max = 27.08 and the minimum is f min = -27.08
Step-by-step explanation:
to find the maximum and minimum value of the function
f(x,y) = 3*x + y
on the region inside the ellipse g(x,y) = x²+36*y²=1
we can use Lagrange multipliers method defining
F(x,y) = f(x,y) - λ* g(x,y) , where g(x,y) = x²+36*y² - 1
such that
Fx(x,y) = fx(x,y) - λ* gx(x,y) = 0
Fy(x,y) = fy(x,y) - λ* gy(x,y) = 0
g(x,y)=0
where the subindices x and y represent the partial derivatives with respect to x and y
then
fx(x,y) - λ* gx(x,y) = 3 - λ*(2*x) = 0 → x =3/(2*λ)
fy(x,y) - λ* gy(x,y) = 1 - λ*(36*2*y) = 0 → y =1/(72*λ)
x²+36*y² - 1 = 0 → [3/(2*λ)]²+36*[1/(72*λ)]² - 1 = 0
9/(4*λ²) + 1/(144*λ²) - 1 = 0
1/λ²* ( 9/4 + 1/144) = 1
λ = ±√(9/4 + 1/144) = ±√(9/4 + 1/144) = ±(5/12) *√13
since
f(x,y) = 9/(2*λ) + 1/(72*λ)
f max inside = 9/(2*λ) + 1/(72*λ) = 9*6/√13 +6/(5*√13)= 15.3
f min inside = -15.3
for the boundary
f(x,y) = 3*x + y
and x²+36*y² = 1 , differentiating with respect to x
2*x + 72*y*dy/dx = 0
dy/dx = -x/(36*y)
and
df/dx = 3 + dy/dx =0
3 - x/(36*y) = 0
x= 108*y → x²+36*y² = 1 → 108²*y²+36*y² = 1
→ y= ±1/12 → x= 108*y= ±9
then
f max boundary = 3*9 + 1/12 = 27.08
f min boundary = -27.08
therefore comparing the maximum and minimum values inside the ellipse and in the boundary we can conclude
f max = 27.08
f min = -27.08
