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Average rate of change of cost on [10, 15]

A company's total cost (in dollars) to produce xx units of a product is given by C(x)=0.5x2+20x+100.C(x) = 0.5x^2 + 20x + 100.

Find the average rate of change of C(x)C(x) over the interval [10,15][10, 15]. This quantity represents the average cost per additional unit produced as production increases from 10 to 15 units. What is its value?

Hint

Use the secant slope formula C(15)−C(10)15−10\dfrac{C(15) - C(10)}{15 - 10}.

Step-by-step solution

First compute C(10)=0.5(10)2+20(10)+100=50+200+100=350C(10) = 0.5(10)^2 + 20(10) + 100 = 50 + 200 + 100 = 350. Then compute C(15)=0.5(15)2+20(15)+100=112.5+300+100=512.5C(15) = 0.5(15)^2 + 20(15) + 100 = 112.5 + 300 + 100 = 512.5.

The average rate of change is C(15)−C(10)15−10=512.5−3505=162.55=32.5\dfrac{C(15) - C(10)}{15 - 10} = \dfrac{512.5 - 350}{5} = \dfrac{162.5}{5} = 32.5. This is the slope of the secant line through the points (10,350)(10, 350) and (15,512.5)(15, 512.5).

Instantaneous marginal cost at x = 10

Using the limit definition of the derivative, C′(a)=lim⁡x→aC(x)−C(a)x−a,C'(a) = \lim_{x \to a} \frac{C(x) - C(a)}{x - a}, find the instantaneous rate of change of cost at x=10x = 10 (i.e., C′(10)C'(10)). This value approximates the cost of producing one more unit beyond 10 units.

Hint

Set a=10a = 10, write C(x)−C(10)C(x) - C(10) as a polynomial in xx, factor out (x−10)(x-10) from the numerator, then take the limit as x→10x \to 10.

Step-by-step solution

We have C(x)−C(10)=0.5x2+20x+100−350=0.5x2+20x−250C(x) - C(10) = 0.5x^2 + 20x + 100 - 350 = 0.5x^2 + 20x - 250. Factoring, 0.5x2+20x−250=0.5(x2+40x−500)=0.5(x−10)(x+50)0.5x^2 + 20x - 250 = 0.5(x^2 + 40x - 500) = 0.5(x - 10)(x + 50).

So C(x)−C(10)x−10=0.5(x+50)\dfrac{C(x) - C(10)}{x - 10} = 0.5(x + 50) for x≠10x \neq 10. Taking the limit as x→10x \to 10 gives C′(10)=0.5(10+50)=0.5(60)=30C'(10) = 0.5(10 + 50) = 0.5(60) = 30.

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