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The following function gives the cost, in dollars, of producing 
x gallons of wood stain:

C(x)=0.0004x^(3)-0.001x^(2)+0.1 x+3200
What is the instantaneous rate of change of the cost when 100 gallons are produced?
Choose 1 answer:
(A) 11.9 dollars
(B) 11.9 dollars per gallon
(C) 3600 dollars
(D) 3600 dollars per gallon

The following function gives the cost, in dollars, of producing x x gallons of wood stain:\newlineC(x)=0.0004x30.001x2+0.1x+3200 C(x)=0.0004 x^{3}-0.001 x^{2}+0.1 x+3200 \newlineWhat is the instantaneous rate of change of the cost when 100100 gallons are produced?\newlineChoose 11 answer:\newline(A) 1111.99 dollars\newline(B) 1111.99 dollars per gallon\newline(C) 36003600 dollars\newline(D) 36003600 dollars per gallon

Full solution

Q. The following function gives the cost, in dollars, of producing x x gallons of wood stain:\newlineC(x)=0.0004x30.001x2+0.1x+3200 C(x)=0.0004 x^{3}-0.001 x^{2}+0.1 x+3200 \newlineWhat is the instantaneous rate of change of the cost when 100100 gallons are produced?\newlineChoose 11 answer:\newline(A) 1111.99 dollars\newline(B) 1111.99 dollars per gallon\newline(C) 36003600 dollars\newline(D) 36003600 dollars per gallon
  1. Calculate Derivative: To find the instantaneous rate of change, we need to calculate the derivative of the cost function C(x)C(x) with respect to xx.
  2. Find C(x)C'(x): Differentiate C(x)=0.0004x30.001x2+0.1x+3200C(x) = 0.0004x^3 - 0.001x^2 + 0.1x + 3200 with respect to xx to get C(x)C'(x).\newlineC(x)=0.0012x20.002x+0.1C'(x) = 0.0012x^2 - 0.002x + 0.1
  3. Substitute x=100x=100: Plug in x=100x = 100 into C(x)C'(x) to find the rate of change at that point.\newlineC(100)=0.0012(100)20.002(100)+0.1C'(100) = 0.0012(100)^2 - 0.002(100) + 0.1
  4. Calculate C(100)C'(100): Calculate the value of C(100)C'(100).
    C(100)=0.0012(10000)0.002(100)+0.1C'(100) = 0.0012(10000) - 0.002(100) + 0.1
    C(100)=120.2+0.1C'(100) = 12 - 0.2 + 0.1
    C(100)=11.9C'(100) = 11.9

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