Chapter 14.5, Problem 4CP

Mathematical Applications for the ...

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

Chapter
Section

Mathematical Applications for the ...

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

Find the minimum value of f ( x ,   y ) =   x 2 + y 2 -  4 x y , subject to the constraint  x   +   y   = 10 ,  by finding the value of f(x, y) at the critical values of x and y.

To determine

To calculate: The minimum value of the function f(x,y)=x2+y24xy subject to the constraint x+y=10 by finding the value of f(x,y) at the critical values of x and y.

Explanation

Given Information:

The provided function is f(x,y)=x2+y2âˆ’4xy subject to the constraint x+y=10.

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 Î» to form the objective function F(x,y,Î»)=f(x,y)+Î»g(x,y).

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

(3) The critical points of F(x,y,Î») are the points that satisfy âˆ‚Fâˆ‚x=0, âˆ‚Fâˆ‚y=0, and âˆ‚Fâˆ‚Î»=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 fâ€²(x)=nxnâˆ’1.

Constant function rule for a constant c is such that, if f(x)=c then fâ€²(x)=0.

Coefficient rule for a constant c is such that, if f(x)=câ‹…u(x), where u(x) is a differentiable function of x, then fâ€²(x)=câ‹…uâ€²(x).

Calculation:

Consider the function, f(x,y)=x2+y2âˆ’4xy.

The provided constraint is x+y=10.

According to the Lagrange multipliers method,

The objective function is F(x,y,Î»)=f(x,y)+Î»g(x,y).

Thus, f(x,y)=x2+y2âˆ’4xy and g(x,y)=x+yâˆ’10.

Substitute x2+y2âˆ’4xy for f(x,y) and x+yâˆ’10 for g(x,y) in F(x,y,Î»)=f(x,y)+Î»g(x,y).

F(x,y,Î»)=x2+y2âˆ’4xy+Î»(x+yâˆ’10)

Therefore, the objective function of f(x,y)=x2+y2âˆ’4xy subject to the constraint x+y=10 is F(x,y,Î»)=x2+y2âˆ’4xy+Î»(x+yâˆ’10)

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