The problem gives the total cost function for a product as $C(x) = 850\ln(x+10) + 1700$, where $x$ is the number of units produced. (a) We need to find the total cost of producing 300 units and round the answer to the nearest cent. (b) We need to find how many units will give a total cost of $8500 and round the answer to the nearest whole number.

Applied MathematicsCost FunctionLogarithmsCalculusOptimizationModeling
2025/3/11

1. Problem Description

The problem gives the total cost function for a product as C(x)=850ln(x+10)+1700C(x) = 850\ln(x+10) + 1700, where xx is the number of units produced.
(a) We need to find the total cost of producing 300 units and round the answer to the nearest cent.
(b) We need to find how many units will give a total cost of $8500 and round the answer to the nearest whole number.

2. Solution Steps

(a) To find the total cost of producing 300 units, we substitute x=300x = 300 into the cost function:
C(300)=850ln(300+10)+1700C(300) = 850\ln(300+10) + 1700
C(300)=850ln(310)+1700C(300) = 850\ln(310) + 1700
Using a calculator, we find that ln(310)5.73669\ln(310) \approx 5.73669.
C(300)=850(5.73669)+1700C(300) = 850(5.73669) + 1700
C(300)=4876.1865+1700C(300) = 4876.1865 + 1700
C(300)=6576.1865C(300) = 6576.1865
Rounding to the nearest cent, we get 6576.196576.19.
(b) To find the number of units that will give a total cost of 8500,weset8500, we set C(x) = 8500andsolvefor and solve for x$:
8500=850ln(x+10)+17008500 = 850\ln(x+10) + 1700
Subtract 1700 from both sides:
85001700=850ln(x+10)8500 - 1700 = 850\ln(x+10)
6800=850ln(x+10)6800 = 850\ln(x+10)
Divide both sides by 850:
6800850=ln(x+10)\frac{6800}{850} = \ln(x+10)
8=ln(x+10)8 = \ln(x+10)
Exponentiate both sides with base ee:
e8=x+10e^8 = x+10
x=e810x = e^8 - 10
Using a calculator, we find that e82980.957987e^8 \approx 2980.957987.
x=2980.95798710x = 2980.957987 - 10
x=2970.957987x = 2970.957987
Rounding to the nearest whole number, we get 29712971.

3. Final Answer

(a) $6576.19
(b) 29712971

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