The function f(x, y) = xy has an absolute maximum value and absolute minimum value subject to the constraint 2x^2 + 3y^2 - 3xy = 49. Use Lagrange multipliers to find these values. A. Find the gradient of f(x, y) = 3xy.
B. Find the gradient of g(x, y) = x2 + y2 - xy-9.
C. Write the Lagrange multiplier conditions.
A. 3xy = M2x - y), 3xy = M2y - x), x2 + y2 - xy - 9 = 0
B. 3x = M(2x - y), 3y = M2y - x), 3xy = 0
C. 3y = M(2x - y), 3x = M(2y – x), x² + y2 - xy - 9 = 0
D. 3x = M(2x - y), 3y = M2y - x), x2 + y2 - xy - 9 = 0
The absolute maximum value is:.
The absolute minimum value is:.

The absolute minimum of f(x,y) = 8.107

The absolute maximum of f(x,y) = 24.326

Step-by-step explanation:

f(x,y) = xy. The constraint equation is 2x² + 3y² - 3xy = 49

Let g(x,y) = 2x² + 3y² - 3xy - 49

df/dx = y and df/dy = x , dg/dx = 4x - 3y and dg/dy = 6y - 3x

Using Lagrange multipliers,

df/dx = λdg/dx and df/dy = λdg/dy

So,

y = λ(4x - 3y)   (1 )and x = λ(6y - 3x)  (2)

y = 4λx - 3λy

y + 3λy = 4λx

y(1 + 3λ) = 4λx

y = 4λx/(1 + 3λ)

Substituting y into (2), we have

x = λ(6y - 3x)

x = λ(6[4λx/(1 + 3λ)] - 3x)

x =  24λ²x/(1 + 3λ) - 3λx

24λ²x/(1 + 3λ) - 3λx - x = 0

[24λ²/(1 + 3λ) - 3λ - 1]x = 0

⇒ [24λ²/(1 + 3λ) - 3λ - 1] = 0 since x ≠ 0

[24λ²/(1 + 3λ) - 3λ - 1] = 0

⇒[24λ²/(1 + 3λ) - (3λ + 1)] = 0

[24λ² - (3λ + 1)²] = 0

24λ² - 9λ² - 6λ - 1 = 0

15λ² - 6λ - 1 = 0

Using the quadratic formula,

λ =

λ = (6 + 4√6)/30 or (6 - 4√6)/30

λ = (3 + 2√6)/15 = 0.527 or (3 - 2√6)/15 = -0.127

Substituting y into the constraint equation, we have

2x² + 3y² - 3xy = 49

2x² + 3(4λx/(1 + 3λ))² - 3x(4λx/(1 + 3λ)) = 49

2x² + 12λ²x²/(1 + 3λ))² - 12λx²/(1 + 3λ) = 49

[2 + 12λ²/(1 + 3λ)² - 12λ/(1 + 3λ)}x² = 49

[2(1 + 3λ)² + 12λ² - 12λ(1 + 3λ)]x²/(1 + 3λ)² = 49

[2(1 + 6λ + 9λ²) + 12λ² - 12λ + 36λ²)]x²/(1 + 3λ)² = 49

[2 + 12λ + 18λ² + 12λ² - 12λ + 36λ²)]x²/(1 + 3λ)² = 49

[2 + 6λ²]x²/(1 + 3λ)² = 49

x² = 49(1 + 3λ)²/(2 + 6λ²)

x² = 49(1 + 3λ)²/2(1 + 3λ²)

x = √[49(1 + 3λ)²/2(1 + 3λ²)]

x = ±7√[(1 + 3λ)²/2(1 + 3λ²)]

Substituting λ = (3 + 2√6)/15 = 0.527 or (3 - 2√6)/15 = -0.127

x = ±7√[(1 + 3(0.527))²/2(1 + 3(0.527)²)] or ±7√[(1 + 3(-0.127))²/2(1 + 3(-0.127)²)]

x = ±7√[(6.662/3.666] or ±7√[0.3831/1.9032)]

x = ±7√1.8172 or ±7√0.2012

x = ±9.436 or ±3.141

Substituting x and λ into y, we have

y = 4λx/(1 + 3λ)

y = 4(0.527)(±3.141)/(1 + 3(0.527)) or  4(-0.127)(±3.141)/(1 + 3(-0.127))

y = ±6.6621/2.581    or ±1.5956/0.619

y = ±2.581 or ±2.578

The minimum value of f(x,y) is gotten at the minimum values of x and y which are x = -3.141 and y = -2.581

So f(-3.141,-2.581) = -3.141 × -2.581 = 8.107

The maximum value of f(x,y) is gotten at the minimum values of x and y which are x = +9.436 and y = +2.578

So f(+9.436,+2.578) = +9.436 × +2.578 = 24.326

the answer is d. (1.2, -4.7) on edgn

hope this !

answer: yes, since the line has a steeper rate of change than the exponential the y-values will be greater and they will only intersect one time.