Find the maximum and minimum values of f (x, y) = 3x+yf (x, y) = 3x+y on the ellipse x2+36y2=1x2+36y2=1 maximum value: minimum value:

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