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

answer is the third choice:

3x^5+4x^4+x^3+2x^2-10x-7=0

step-by-step explanation: