Math Problem Statement

Use the method of Lagrange multipliers to find the minimum value of 𝑓(π‘₯, 𝑦) = 6π‘₯ 2 + 5𝑦 2 βˆ’ π‘₯𝑦 subject to the constraint 2π‘₯ + 𝑦 = 24

Solution

To find the minimum value of f(x,y)=6x2+5y2βˆ’xyf(x, y) = 6x^2 + 5y^2 - xy subject to the constraint 2x+y=242x + y = 24, we will use the method of Lagrange multipliers. The steps are as follows:

  1. Define the Lagrange function:

    L(x,y,Ξ»)=f(x,y)+Ξ»(g(x,y)βˆ’c)\mathcal{L}(x, y, \lambda) = f(x, y) + \lambda (g(x, y) - c)

    where f(x,y)=6x2+5y2βˆ’xyf(x, y) = 6x^2 + 5y^2 - xy and the constraint g(x,y)=2x+yβˆ’24g(x, y) = 2x + y - 24.

    Therefore, the Lagrange function is:

    L(x,y,Ξ»)=6x2+5y2βˆ’xy+Ξ»(2x+yβˆ’24)\mathcal{L}(x, y, \lambda) = 6x^2 + 5y^2 - xy + \lambda (2x + y - 24)

  2. Compute the partial derivatives and set them equal to zero:

    βˆ‚Lβˆ‚x=12xβˆ’y+2Ξ»=0\frac{\partial \mathcal{L}}{\partial x} = 12x - y + 2\lambda = 0 βˆ‚Lβˆ‚y=10yβˆ’x+Ξ»=0\frac{\partial \mathcal{L}}{\partial y} = 10y - x + \lambda = 0 βˆ‚Lβˆ‚Ξ»=2x+yβˆ’24=0\frac{\partial \mathcal{L}}{\partial \lambda} = 2x + y - 24 = 0

  3. Solve the system of equations:

    From the third equation:

    y=24βˆ’2xy = 24 - 2x

    Substitute y=24βˆ’2xy = 24 - 2x into the first and second equations:

    12xβˆ’(24βˆ’2x)+2Ξ»=012x - (24 - 2x) + 2\lambda = 0 10(24βˆ’2x)βˆ’x+Ξ»=010(24 - 2x) - x + \lambda = 0

    Simplify these equations:

    12xβˆ’24+2x+2Ξ»=0β€…β€ŠβŸΉβ€…β€Š14xβˆ’24+2Ξ»=0β€…β€ŠβŸΉβ€…β€Š2Ξ»=24βˆ’14xβ€…β€ŠβŸΉβ€…β€ŠΞ»=12βˆ’7x12x - 24 + 2x + 2\lambda = 0 \implies 14x - 24 + 2\lambda = 0 \implies 2\lambda = 24 - 14x \implies \lambda = 12 - 7x

    240βˆ’20xβˆ’x+Ξ»=0β€…β€ŠβŸΉβ€…β€Š240βˆ’21x+Ξ»=0β€…β€ŠβŸΉβ€…β€ŠΞ»=21xβˆ’240240 - 20x - x + \lambda = 0 \implies 240 - 21x + \lambda = 0 \implies \lambda = 21x - 240

    Set the two expressions for Ξ»\lambda equal to each other:

    12βˆ’7x=21xβˆ’240β€…β€ŠβŸΉβ€…β€Š252=28xβ€…β€ŠβŸΉβ€…β€Šx=912 - 7x = 21x - 240 \implies 252 = 28x \implies x = 9

    Substitute x=9x = 9 back into the constraint equation to find yy:

    y=24βˆ’2(9)=6y = 24 - 2(9) = 6

  4. Verify the solution:

    Substitute x=9x = 9 and y=6y = 6 back into the original function f(x,y)f(x, y):

    f(9,6)=6(92)+5(62)βˆ’(9β‹…6)f(9, 6) = 6(9^2) + 5(6^2) - (9 \cdot 6) f(9,6)=6(81)+5(36)βˆ’54f(9, 6) = 6(81) + 5(36) - 54 f(9,6)=486+180βˆ’54f(9, 6) = 486 + 180 - 54 f(9,6)=612f(9, 6) = 612

Therefore, the minimum value of the function f(x,y)=6x2+5y2βˆ’xyf(x, y) = 6x^2 + 5y^2 - xy subject to the constraint 2x+y=242x + y = 24 is 612.

Would you like any more details or have any questions?

Here are some related questions you might find interesting:

  1. How do Lagrange multipliers work in three variables?
  2. Can Lagrange multipliers be used for inequality constraints?
  3. How can we verify if the critical point found is a minimum or maximum?
  4. What are some real-life applications of Lagrange multipliers?
  5. How does the method of Lagrange multipliers compare to other optimization methods?

Tip: When dealing with optimization problems, it's crucial to carefully check the constraint conditions and ensure that the critical points satisfy these constraints.

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Math Problem Analysis

Mathematical Concepts

Optimization
Lagrange multipliers

Formulas

Lagrange function

Theorems

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Suitable Grade Level

Advanced undergraduate level