   Chapter 14.5, Problem 3E Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

Solutions

Chapter
Section Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

Find the minimum value of z   =   3 x 2 + 5 y 2 −   2 x y subject to the constraint x   +   y   =   5 .

To determine

To calculate: The minimum value of z=3x2+5y22xy subject to the constraint x+y=5.

Explanation

Given Information:

The provided function is z=3x2+5y22xy subject to the constraint x+y=5.

Formula used:

According to the Lagrange multipliers method to obtain maxima or minima for a function z=f(x,y) subject to the constraint g(x,y)=0,

(1) Find the critical values of f(x,y) using the new variable λ to form the objective function F(x,y,λ)=f(x,y)+λg(x,y).

(2) The critical points of f(x,y) are the critical values of F(x,y,λ) which satisfies g(x,y)=0.

(3) The critical points of F(x,y,λ) are the points that satisfy Fx=0, Fy=0, and Fλ=0, that is, the points which make all the partial derivatives of zero.

For a function f(x,y), the partial derivative of f with respect to y is calculated by taking the derivative of f(x,y) with respect to y and keeping the other variable x constant. The partial derivative of f with respect to y is denoted by fy.

Power of x rule for a real number n is such that, if f(x)=xn then f(x)=nxn1.

Constant function rule for a constant c is such that, if f(x)=c then f(x)=0.

Coefficient rule for a constant c is such that, if f(x)=cu(x), where u(x) is a differentiable function of x, then f(x)=cu(x).

Calculation:

Consider the function, z=3x2+5y22xy.

The provided constraint is x+y=5.

According to the Lagrange multipliers method,

The objective function is F(x,y,λ)=f(x,y)+λg(x,y).

Thus, f(x,y)=3x2+5y22xy and g(x,y)=x+y5.

Substitute 3x2+5y22xy for f(x,y) and x+y5 for g(x,y) in F(x,y,λ)=f(x,y)+λg(x,y).

F(x,y,λ)=3x2+5y22xy+λ(x+y5)

Since, the critical points of F(x,y,λ) are the points that satisfy Fx=0, Fy=0, and Fλ=0

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