Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. f(x,y)= e^xy; X^3+y^3=16

Question

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. f(x,y)= e^xy; X^3+y^3=16

in progress 0
Diễm Thu 5 years 2021-08-30T07:48:11+00:00 1 Answers 10 views 0

Answers ( )

  1. trucchi
    0
    2021-08-30T07:49:15+00:00
    Reply

    Answer:

    f(x,y) = e^{xy} is maximum at x = 2 and y = 2 and f(2,2) = e^{4}

    Step-by-step explanation:

    Since f(x,y) = e^{xy} and x³ + y³ = 16, Ф(x,y) = x³ + y³ – 16

    df/dx = ye^{xy}, df/dy = xe^{xy}, dФ/dx = 3x² and dФ/dy = 3y²

    From the method of Lagrange multipliers,

    df/dx = λdΦ/dx and df/dy = λdΦ/dy

    ye^{xy} = 3λx² (1) and  xe^{xy} = 3λy² (2)

    multiplying (1) by x and (2) by y, we have

    xye^{xy} = 3λx³ (4) and  xye^{xy} = 3λy³ (5)

    So,  3λx³ = 3λy³

    ⇒ x = y

    Substituting x = y into the constraint equation, we have

    x³ + y³ = 16

    x³ + x³ = 16

    2x³ = 16

    x³ = 16/2

    x³ = 8

    x = ∛8

    x = 2 ⇒ y = 2, since x = y

    So, f(x,y) = f(2,2) = e^{2 X2} = e^{4}

    We need to determine if this is a maximum or minimum point by considering other points that fit into the constraint equation.

    Since x³ + y³ = 16 when x = 0, y is maximum when y = 0, x = maximum

    So, 0³ + y³ = 16

    y³ = 16

    y = ∛16

    Also, when y = 0, x = maximum

    So, x³ + 0³ = 16

    x³ = 16

    x = ∛16

    and f(0,∛16) = e^{0X\sqrt[3]{16} } = e^{0} = 1.

    Also, f(∛16, 0) = e^{\sqrt[3]{16}X0 } = e^{0} = 1.

    Since f(0,∛16) = f(∛16, 0) = 1 < f(2,2) = e^{4}

    f(2,2) is a maximum point

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )