{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Maxima Minima

Maxima Minima - Contents 8 Applications of Maxima and...

This preview shows pages 1–5. Sign up to view the full content.

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

View Full Document

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Contents 8 Applications of Maxima and Minima 127 8.1 The Use of Auxiliary Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 2 CONTENTS Chapter 8 Applications of Maxima and Minima Prerequisites When you write a solution, it is your responsibility to inform the reader what you are doing. You may assume the marker is intelligent but not a mind reader. Therefore you must invest some time in the organization and documentation of your work. Study the following problems and solution with these points in mind. Problem A lighthouse is at point P , 3 miles offshore, from the nearest point 0 of a straight beach. A store is located 5 miles down the beach from 0. The lighthouse keeper can row 3.25 miles/hour and walk 4 miles/hour. How far along the beach from 0 should he land in order to get to the store in the minimum time? Solution: # # # # # # # # # # # # # 5- x 3 √ x 2 + 9 x X P S Let x be the distance from 0 to the point of landing X and let S be the location of the store 5 miles from 0. A. Compute the time function for the trip: The time required to row from P to X is √ x 2 +9 3 . 25 . (Recall that velocity = distance/time). The time required to walk from X to S is 5- x 4 . The total time, T , for the trip is given by: T ( x ) = √ x 2 + 9 3 . 25 + 5- x 4 . In this problem it only makes sense for x to vary from 0 to 5 miles. Therefore we need to find the minimum of T on the closed interval [0 , 5]. 128 Applications of Maxima and Minima B. Compute the critical points of the function: dT dx = x 3 . 25 √ x 2 + 9- 1 4 . Therefore dT dx = ⇔ 4 x = 3 . 25 p x 2 + 9 ⇔ 16 x 2 = (3 . 25) 2 ( x 2 + 9) ⇔ { 16- (3 . 25) 2 } x 2 = 9(3 . 25) 2 ⇔ x 2 ∼ = 17 . 482759 ⇔ x ∼ = 4 . 1812389 or- 4 . 1812389 C. Evaluate the function at the critical and endpoints to find the value of x that minimizes the trip. T (0) ∼ = 2 . 1730769 T (5) ∼ = 1 . 794139 T (4 . 1812389) ∼ = 1 . 788118 D. Conclusion. Choose x = 4 . 1812389 to minimize the time of the trip. Observe that a very slight increase in rowing speed would make rowing directly to S the fastest route. The following problems may be done by using a four step technique which is outlined below. Step 1 (a) Draw a diagram. The diagram should be large and neatly drawn, occupying about 1 3 page. (b) Label fixed quantities with numerical values. Label the dimensions that vary with letters. Choose letters for any other items that are involved in the problem, e.g. volume. Step 2 (a) Select the quantity that is to be made maximum or minimum. Express it as a function of other quantities. (b) If you are to maximize something which is a function of two variables, then you must find a relationship between the two variables. Usually the geometry of the situation gives a relation. (e.g. similar triangles, Pythagoras theorem.) (c) Express the quantity to be maximized as a function of only one variable....
View Full Document

{[ snackBarMessage ]}

Page1 / 22

Maxima Minima - Contents 8 Applications of Maxima and...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online