   Chapter 7.6, Problem 24E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Finding Distance In Exercises 23-28, find the minimum distance from the curve or surface to the given point. (Hint: Start by minimizing the square of the distance.)Line: 2 x + 3 y = − 1 , ( 0 , 0 ) Minimize d 2 = x 2 + y 2

To determine

To calculate: The minimum distance from the line 2x+3y=1, to the point (0,0), where the square of the distance d2=x2+y2

Explanation

Given Information:

The provided distance is d2=x2+y2 and constraint is 2x+3y+1=0.

Formula used:

Method of Lagrange multipliers,

If the function f(x,y) contains a maximum or minimum subject to the constraint g(x,y)=0 then the maximum or minimum can occur at one of the critical numbers of the function F is,

F(x,y,λ)=f(x,y)λg(x,y) where, λ is a Lagrange multiplier.

Steps to determine the minimum or maximum of the function f.

1. Solve the system of equations,

Fx(x,y,λ)=0Fy(x,y,λ)=0Fλ(x,y,λ)=0

2. Determine the value of the function f at each solution obtained from the step 1.

The largest value gives the maximum value of function f subject to the constraint g(x,y)=0 and the lowest value gives the minimum value of function f subject to the constraint g(x,y)=0

Calculation:

Consider the function,

d2=x2+y2=f(x,y)

The provided constraint is 2x+3y+1=0.

So, g(x,y)=2x+3y+1

For the minimum distance the function f is defined as,

F(x,y,λ)=x2+y2λ(2x+3y+1)

Now, differentiate with respect to x, y and λ, and then set each partial derivative equal to zero as,

Fx(x,y,λ)=02x2λ=0λ=x

And,

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