Answered: Find the closest and most distant…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter9: Systems Of Equations And Inequalities

Section: Chapter Questions

Problem 39RE

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Question

Find the closest and most distant points to the origin on the curve x^2+xy+y^2=1. Use lagrange multipliers and identify the function and the constraint,

Expert Solution

Step 1

Given that the curve is $x^{2} + x y + y^{2} = 1$ .

It is required to find the closest and most distant points to the origin on the curve $x^{2} + x y + y^{2} = 1$ .Thus the function is $f (x, y) = x^{2} + x y + y^{2} - 1 = 0$ .

Sketch the graph of the curve and obtain the constraint.

The constraint is that the points are from origin on the curve $x^{2} + x y + y^{2} = 1$ .That is the constraints become $g (x, y) = x^{2} + y^{2}$

Step 2

First find $\nabla f (x, y) a n d \nabla g (x, y)$ .

$\nabla f (x, y) = <f_{x}, f_{y}> = <2 x + y, x + 2 y> \nabla g (x, y) = <g_{x}, g_{y}> = <2 x, 2 y>$

Now substitute in $\nabla g (x, y) = λ \nabla f (x, y)$ .

$<2 x, 2 y> = λ <2 x + y, x + 2 y> \Rightarrow λ (2 x + y) = 2 x λ = \frac{2 x}{2 x + y} \Rightarrow λ (x + 2 y) = 2 y λ = \frac{2 y}{x + 2 y}$

On further simplification,

$\frac{2 x}{2 x + y} = \frac{2 y}{2 y + x} 2 x (2 y + x) = 2 y (2 x + y) 4 x y + 2 x^{2} = 4 x y + 2 y^{2} 2 x^{2} = 2 y^{2} x^{2} = y^{2} x = \pm y$

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