Applications of Differentiation
(b) Economics Economics insights:
1. Profi t (P ) = Revenue or income (R) – Costs (C ) 2. N objects sold at € x each ⇒ Revenue R = € Nx
EXAMPLE 25
The fuel economy E (km/l) of an average saloon car is given by E = – 0·00375 v 2 + 0·575v − 2, where v is the speed of the car in km/h. At what speed, correct to one decimal place, is the fuel economy a maximum?
Solution 1. Fuel economy (E ) 3. E = − 0·00375 v 2 + 0·575v − 2
6. dE ___
dv = – 0·0075v + 0·575 −0·0075v + 0·575 = 0
v = 76·7 [There is one stationary point.] Fuel economy is a maximum at a speed of 76·7 km/h.
EXAMPLE 26
A company manufactures and sells shirts. The company’s fi xed costs are €1500 per week. The production costs are €10 per shirt. For a selling price of € x per shirt, the number N of shirts sold per week is given by N = −10x + 1200. If the company sells the shirts at € x each, fi nd:
(a) the weekly revenue in terms of x, (b) the weekly costs in terms of x, (c) the weekly profi t in terms of x, (d) the maximum weekly profi t.
Solution
(a) For a selling price of € x per shirt, the number sold is N.
Revenue R per week = Nx = x (−10x + 1200) = 1200x − 10 x 2
(c) Algebraic
The sum of two numbers is 32. Find the numbers if their product is to have its maximum value. 27
EXAMPLE Solution
1. P ( product) 2. Let the numbers be x and y. 3. P = xy 4. x + y = 32 y = (32 − x)
5. P = x(32 − x) = 32x − x 2 6. dP
___ dx = 32 − 2x
32 − 2x = 0 x = 16 [There is one stationary point.]
y = 16 331
(b) Weekly costs: C = Fixed costs + Cost per shirt × Number of shirts C = 1500 + 10(−10x + 1200)
C = −100x + 13 500
(c) P = R − C = 1200x − 10 x 2 − (13 500 − 100x) = −10 x 2 + 1300x − 13 500
Now, for the optimisation of the profi t P.
(d) P = −10 x 2 + 1300x − 13 500 dP
___ dx = −20x + 1300
___ dx = 0 −20x + 1300 = 0
dP
x = 65 [There is only one stationary point.] Maximum profi t: Pmax = −10(65 ) 2 + 1300(65) − 13 500 = €28 750
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