P (x) = 0 6(x+2)(x 3) = 0

Transcription

1 Math Assignment 6 Solutions - Spring Jaimos F Skriletz 1 1. Polynomial Functions Consider the polynomial function P(x) = x 3 6x 18x+16. First Derivative - Increasing, Decreasing, Local Extrema The critical points are found by solving P (x) = 6x 1x 18 = 6(x x 3) = 6(x+)(x 3) P (x) = 0 6(x+)(x 3) = 0 Thus the two solutions (or critical points) are x = and x = 3. Use these critical points to create the following sign table for P (x): From this it follows that P(x) is increasing on (, ). P(x) is decreasing on (,3). P(x) is increasing on (3, ). 3 P (x) (+) ( ) (+) At x =, P(x) goes from increasing to decreasing so the point (,P( )) = (,1) is a local maximum. At x = 3, P(x) goes from decreasing to increasing so the point (3,P(3)) = (3, 38) is a local minimum. Second Derivative - Concavity P (x) = 1x 1 = 1(x 1) The possible inflection points are found by solving the equation P (x) = 0 1(x 1) = 0 Thus the only possible inflection point is at x = 1. Use this to create the following sign table for P (x): P(x) is concave down on (,1). P(x) is concave up on (1, ). 1 P (x) ( ) (+) At x = 1, P(x) changes concavity so the point (1,P(1)) = (1, 6) is an inflection point. Graph The y-intercept of P(x) is (0,P(0)) = (0,16). The graph of P(x) is:

2 Math Assignment 6 Solutions - Spring Jaimos F Skriletz. Rational Functions Consider the rational function R(x) = x 4x+8 x 4x 5 : End Behavior - Horizontal Asymptote The end behavior (x ± ) of R(x) is Thus R(x) has a horizontal asymptote at y = 1. Domain - Vertical Asymptote The domain of R(x) is all real x such that x 4x+8 x ± x 4x 5 = x ( 1 4 x + ) 8 x x ± x ( 1 4 x ) 5 = = 1 x x 4x 5 0 (x+1)(x 5) 0 Thus the domain is {x : x 1 and x 5}. Looking at the it of each of these points shows that 13 R(x) = = ± and x 1 0 R(x) = 13 x 5 0 = ± Thus by it forms we see that R(x) has vertical asymptotes at x = 1 and x = 5. Zeros - x-intercepts The zeros of R(x) are R(x) = 0 x 4x+8 = 0 Since the discriminate of the above quadratic equation is b 4ac = ( 4) 4(1)(8) = 16 < 0 There are no real solutions, thus no zeros or x-intercepts. Sign Table The domain and zeros split the real numbers up into the sign table: From this we get the behavior about the asymptotes is 1 5 R(x) (+) ( ) (+) R(x) = + x 1 x 1 +R(x) = R(x) = x 5 x 5 R(x) = + Local Extrema The critical points for R(x) are R (x) = (x 4)(x 4x 5) (x 4)(x 4x+8) (x 4x 5) = 6(x ) (x 4x 5) R (x) = 0 6(x ) = 0 The only critical point is x =. The sign table for R (x) is x =

3 Math Assignment 6 Solutions - Spring Jaimos F Skriletz 3 Thus we have the following R(x) is increasing on (, 1) and ( 1,). R(x) is decreasing on (,5) and (5, ). (,R()) = (, 4/9) is a local maximum. Graph 1 5 R (x) (+) (+) ( ) ( ) 10 5 K10 K K5 K10 3. Revenue, Cost, Profit Analysis A movie theater has a maximum occupancy of 3850 people. According to the theaters records, if you charge $6 per ticket you get an average attendance of 800 people, but for every 10-cent increase in the ticket price, 35 less people attend. Along with the ticket price, each customer spends on average $5 on concessions. Let x be the number of customers and p be the ticket price. (a) Take the demand to be linear with slope m = p x = = 1 And passes through the point (x 1,p 1 ) = (800,6). So the demand relation is (b) The total revenue function is p p 1 = m(x x 1 ) p 6 = 1 (x 800) p = 1 x+14 R(x) = (ticket sales)+(concessions) = px+5x = ( ) 1 x+14+5 x = 1 x +19x (c) Suppose the fixed costs are $9,450 and the variable costs are an average of $7 per customer, then the cost function is C(x) = 7x+9450 (d) The profit function is P(x) = R(x) C(x) = 1 x +1x 9450 (e) The domain of interest in this problem is 0 x 3850 (theater size).

4 Math Assignment 6 Solutions - Spring Jaimos F Skriletz 4 (f) The break even points occur when R(x) = C(x) 1 x +19x = 7x+9450 x +6650x = 450x x +400x = 0 (g) By the quadratic formula we get the two solutions (rounded to the nearest person) x = 1050 and x = 3150, which represent the two break even points. The critical point of R(x) is R(x) = 1 x +19x R (x) = 0 x+19 = 0 x = 19 ( ) x = 19 = 335 Since R (x) = < 0, the revenue has a maximal value when x = 335. Further more if x = 335 then p = 4.50, so maximum revenue occurs with a ticket price of $4.50 per ticket. (h) P(x) = 1 x +1x 9450 The critical point of P(x) is P (x) = 0 x+1 = 0 x = 1 ( ) x = 1 = 100 Since P (x) = < 0, the profit as a maximal value when x = 100. Thus maximum profit occurs with a ticket price of p = $8 per ticket. (i) The Graphs of R(x), C(x) and P(x) are 30,000 0,000 10, ,000,000 3,000 Maximum Values Break Even Values C(x) R(x) P(x)

5 Math Assignment 6 Solutions - Spring Jaimos F Skriletz 5 4. Taxation The demand equation for snowboards is p = x dollars, where x is in thousands of snowboards, and the 4 manufacturing cost is given by C(x) = 1 4 x +50x+35, where x is in thousands of snowboards and C is in thousands of dollars. (a) Find P(x). R(x) = xp = x( x) = 530x 3 4 x. So P(x) = R(x) C(x) = x +480x 35 (b) What price should the snowboards be sold at to maximize profit? P(x) is a parabola facing downwards so the maximum value occurs when P (x) = x+480 = 0 or x = 480 = 40. When x = 40, p = (40) = $. (c) If there was a cost increase (or tax) of T dollars per snowboard, what portion of the cost increase should be passed on to the consumers to keep profit maximized? If a cost increase of T dollars per snowboard is added the new cost function will be: This will make the new profit function C new (x) = C(x)+Tx = 1 4 x +(50+T)x+35 P new (x) = P(x) Tx = x +(480 T)x 35 As before the maximum of the new profit occurs at the critical point. So to keep profits maximized we want P new(x) = x+(480 T) = 0 or x = 40 1 T. Thus the new price to keep profits maximized is: p new = (40 1 T) = T = p old T So to keep profits maximized 3 8 = 37.5% of the cost increase is passed onto the consumers. 5. Maximum Area A text book designer has decided that the pages of a text book should have one-inch margins on the top and bottom and half-inch margins on the sides, as shown. Furthermore, it is stipulated that the printed area of the page is 3 square inches. What are the dimensions of a page that minimize its total area? x 1" 0.5 y 0.5 1" Given that the printed area needs to be 3 square inches, we have the constraint that PA = xy = 3 or y = 3 x. Using this constraint, it follows that the area of the page is: The minimum value will occur when: A = (x+1)( 3 x +) = 3+x+ 3 x + = x+ 3 x +34 da dx = 3 x = 0 Thus it follows that x = 3 = 16 or x = 4. If x = 4 then y = 3 4 its total area are 5 x10 [(x+1)x(y +)]. = 8. So the dimensions of the page that minimize