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Exercise 6.1 · Q15

Q.The total cost C(x)C(x) in Rupees associated with the production of xx units of an item is given by C(x)=0.007x3−0.003x2+15x+4000C(x) = 0.007x^3 - 0.003x^2 + 15x + 4000. Find the marginal cost when 1717 units are produced.

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Marginal cost is C′(x)=0.021x2−0.006x+15C'(x)=0.021x^2-0.006x+15. At x=17x=17, C′(17)=6.069−0.102+15=20.967C'(17)=6.069-0.102+15=20.967 Rupees.

Solution

1. Marginal cost is the derivative of total cost.

C(x)=0.007x3−0.003x2+15x+4000.C(x)=0.007x^3-0.003x^2+15x+4000.

2. Differentiate term by term (the constant 40004000 vanishes):

C′(x)=3(0.007)x2−2(0.003)x+15=0.021x2−0.006x+15.C'(x)=3(0.007)x^2-2(0.003)x+15=0.021x^2-0.006x+15.

3. Evaluate at x=17x=17 (using 172=28917^2=289): …

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