This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MATH 106 CALCULUS I FOR BIO. & SOC. SCI. FALL 2011 INSTRUCTOR: JOS ´ E MANUEL G ´ OMEZ Solutions Homework 9. Please show all your work. (1) (10 Points) What is the largest area for a right angle triangle whose hypotenuse is 10 cm long? Solution: Let’s denote the sides of the triangle by x and y . The area of the triangle is A = xy 2 . We want to maximize the function A . Using the Pythagorean theorem we get x 2 + y 2 = 100. Solving for y we obtain y = ± √ 100 − x 2 . Note that y is a length so it is positive. We conclude then that y = √ 100 − x 2 . It follows that A ( x ) = x √ 100 − x 2 2 . Let’s determine next the domain of A . To do so note that x is a length so that x ≥ 0. On the other hand, since the hypotenuse of the triangle is 10 we get x ≤ 10. Thus we want to maximize the function A on the domain [0 , 10]. Let’s find next the critical points. For this we have A ′ ( x ) = 1 2 parenleftbigg √ 100 − x 2 + − x 2 √ 100 − x 2 parenrightbigg = 1 2 parenleftbigg 100 − 2 x 2 √ 100 − x 2 parenrightbigg Note that A ′ ( x ) = 0 if and only if 1 2 parenleftbigg 100 − 2 x 2 √ 100 − x 2 parenrightbigg = 0 and thus 100 − 2 x 2 = 0. It follows that x 2 = 50 hence x = ± √ 50. Since the domain of A is the interval [0 , 10] we get that √ 50 is a critical point. Also note that A ′ ( x ) does not exist when...
View
Full
Document
This note was uploaded on 01/20/2012 for the course CALC AS.110.106 taught by Professor Josegomez during the Fall '11 term at Johns Hopkins.
 Fall '11
 JoseGomez
 Calculus

Click to edit the document details