Use Lagrange multipliers to find the indicated extrema, assuming that x, and y are positive. Minimize f(x, y)=x2−10x+y2−14y+28; Constraint: x+y=14

- The values of x and y that minimize the function, subject to the given constraint are 6 and 8 respectively.

- The minimum value of the function = -44

Step-by-step explanation:

The Lagrange multiploer method finds the optimum value of a multivariable function subjected to a given constraint

It replaces the function with a Lagrange equivalent which is

L(x, y) = F(x, y) - λ C(x, y)

where λ Is the lagrange multiplier which can be a function of x and y

F(x, y) = x² - 10x + y² - 14y + 28

C(x, y) = x + y - 14

L(x, y) = x² - 10x + y² - 14y + 28 - λ (x + y - 14)

We now take the partial derivatives of the Lagrange function with respect to x, y and λ respecrively. Then solving to obtain values of x, y and λ that correspond to the minimum of the function. Since the first partial derivatives are all equal to 0 at minimum point.

(∂L/∂x) = 2x - 10 - λ = 0 (eqn 1)

(∂L/∂y) = 2y - 14 - λ = 0 (eqn 2)

(∂L/∂λ) = x + y - 14 = 0 (eqn 3)

Equating eqn 1 and 2

2x - 10 - λ = 2y - 14 - λ

2x - 10 = 2y - 14

2y = 2x - 10 + 14

2y = 2x + 4

y = x + 2 (eqn *)

Substitute eqn ^ into eqn 3

x + y - 14 = 0

x + x + 2 - 14 = 0

2x - 12 = 0

2x = 12

x = 6

y = x + 2 = 6 + 2 = 8

2x - 10 - λ = 0

12 - 10 - λ = 0

λ = 2

The values of x and y that minimize the function are 6 and 8 respectively.

F(x, y) = x² - 10x + y² - 14y + 28

At minimum point, x = 6, y = 8

F(x, y) = 6² - 10(6) + 8² - 14(8) + 28 = 36 - 60 + 64 - 112 + 28 = -44

Hope this Helps!!!

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204

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answer