Question

Use differentials to approximate the change in profit corresponding to an increase in sales (or production) of one unit. Then compare this with the actual chang in profit. Function x-Value P=-0.2x2 + 200x - 80 X = 40 dp = dollars AP = dollars Need Help? Read it Watch Tak to a Tutor 4. [1/2 Points) DETAILS PREVIOUS ANSWERS LARAPCALC10 3.8.034. MY NOTES PRACTICE ANOTHER The revenue R for a company selling x units is R = 800x - 0.1x? (a) Use differentials to approximate the change in revenue as the sales increase from 2900 units to 3000 units. dR$ 780 X (b) Compare this with the actual change in revenue. AR - $ 21000 Need Help? Takto a Tutor

Answer 1

3 X= 40 11 11 3 0.22 P(41) - P(40) -0.2(41)*+ 200 (41) -30 2:0 (41) -80 - ( -0.2 (40)*+ 200 (40)-80) 0.2 (41)+ 0.2(40) + 200(41) - 200(40) -% +8% cui-403) 200 (41-40) 0.2 ( 4921) + 200 3 1 + - 984:2 200 op ANSWER de -022? -- 3 -02 2 + 200 x - 80 x=40 -0.2.32 • 200 2-40 = -0.649 + 200 os(4) |-760 ANSWER So A = -734.2 dollars de= -760 dollars O = AR R: 800 x - 0.1 x? R (3000) R (2900) 800 (3000) - 0.1(3000) - ( 800(2900) -0.1 (2900)?/ 800 (3000-2900) -0.1(3000)² + 01 (2900)? (3000²-29002) = ooox loo 0:1 - 80000 Oil 540000 80000 $9000 GR = 21,000 for dr : dR dz at X- 3000 dx= 100 > (3000-2900 = 100) 0.1,2x dk - 800 000 (800 0.2 x) 100 Воо 012 (3000)) too 600 loo 20,000 AR = 21,000 $ dR = 20,000 & 200 X IDO Henee