15.8%20Ex%2027%20-%2037

# 15.8%20Ex%2027%20-%2037 - 786 C H A P T E R 15 D I F F E R...

This preview shows pages 1–3. Sign up to view the full content.

786 CHAPTER 15 DIFFERENTIATION IN SEVERAL VARIABLES (ET CHAPTER 14) 27. Let Q be the point on an ellipse closest to a given point P outside the ellipse. It was known to the Greek mathematician Apollonius (third century BCE)that PQ is perpendicular to the tangent at Q (Figure 13). Explain in words why this conclusion is a consequence of the method of Lagrange multipliers. Hint: The circles centered at P are level curves of the function to be minimized. P Q FIGURE 13 SOLUTION Let P = ( x 0 , y 0 ) . The distance d between the point P and a point Q = ( x , y ) on the ellipse is minimum where the square d 2 is minimum (since the square function u 2 is increasing for u 0). Therefore, we want to minimize the function f ( x , y , z ) = ( x x 0 ) 2 + ( y y 0 ) 2 + ( z z 0 ) 2 subject to the constraint g ( x , y ) = x 2 a 2 + y 2 b 2 = 1 The method of Lagrange indicates that the solution Q is the point on the ellipse where f = λ g , that is, the point on the ellipse where the gradients f and g are parallel. Since the gradient is orthogonal to the level curves of the function, g is orthogonal to the ellipse g ( x , y ) = 1, and f is orthogonal to the level curve of f passing through Q . But this level curve is a circle through Q centered at P , hence the parallel vectors g and f are orthogonal to the ellipse and to the circle centered at P respectively. We conclude that the point Q is the point at which the tangent to the ellipse is also the tangent to the circle through Q centered at P . That is, the tangent to the ellipse at Q is perpendicular to the radius of the circle. 28. Antonio has \$5.00 to spend on a lunch consisting of hamburgers (\$1.50 each) and French fries (\$1.00 per order). Antonio’s satisfaction from eating x 1 hamburgers and x 2 orders of French fries is measured by a function U ( x 1 , x 2 ) = x 1 x 2 . How much of each type of food should he purchase to maximize his satisfaction (assume that fractional amounts of each food can be purchased)? Antonio has \$5.00 to spend on the lunch, hence the total cost 1 . 5 x 1 + x 2 must satisfy 1 . 5 x 1 + x 2 = 5 We thus want to maximize the function U ( x 1 , x 2 ) = x 1 x 2 subject to the constraint g ( x , y ) = 1 . 5 x 1 + x 2 = 5 with x 1 > 0, x 2 > 0. Step 1. Write out the Lagrange Equations. The gradient vectors are U = 1 2 D q x 2 x 1 , q x 1 x 2 E and g = h 1 . 5 , 1 i , hence the Lagrange Condition U = g gives the following equations: 1 2 ± x 2 x 1 = 1 . 5 1 2 ± x 1 x 2 = x 2 x 1 = 9 2 x 1 x 2 = 4 2 Step 2. Solve for x 1 and x 2 using the constraint. The two equations in step 1 give 2 = x 2 9 x 1 = x 1 4 x 2 Therefore, 4 x 2 2 = 9 x 2 1 x 2 2 = 9 4 x 2 1 x 2 = 3 2 x 1 We now substitute x 2 = 3 2 x 1 in the constraint 1 . 5 x 1 + x 2 = 5 and solve for x 1 .Weget 1 . 5 x 1 + 3 2 x 1 = 5 3 x 1 = 5 x 1 = 5 3

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
SECTION 15.8 Lagrange Multipliers: Optimizing with a Constraint (ET Section 14.8) 787 We fnd x 2 by the relation x 2 = 3 2 x 1 : x 2 = 3 2 · 5 3 = 5 2 We obtain the critical point: µ 5 3 , 5 2 Step 3. Conclusions. We conclude that Antonio should have 5 3 hamburgers and 5 2 orders oF Fries, to maximize his sat- isFaction. Notice that U ( x 1 , x 2 ) = x 1 x 2
This is the end of the preview. Sign up to access the rest of the document.

## 15.8%20Ex%2027%20-%2037 - 786 C H A P T E R 15 D I F F E R...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online