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total increase in cost.
Ques: 3. The additional cost (in lakhs of rupees) of producing a notor car is given by
6+4x^(2)+1.5e^(-x), where
x is the quartity produced. Determin: the total cost producing 5 motor cars if the fixed cost is Rs. 7 lakhs. It is given that
e^(-5^(2))=0.006.

total increase in cost. $\newline$ Ques: $3$ . The additional cost (in lakhs of rupees) of producing a notor car is given by $6+4 \mathrm{x}^{2}+1.5 \mathrm{e}^{-\mathrm{x}}$ , where $\mathrm{x}$ is the quartity produced. Determin: the total cost producing $5$ motor cars if the fixed cost is Rs. $7$ lakhs. It is given that $\mathrm{e}^{-5^{2}}=0.006$ .

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Finance problems

Full solution

Q. total increase in cost.

\newline

Ques:

3

. The additional cost (in lakhs of rupees) of producing a notor car is given by

6+4 \mathrm{x}^{2}+1.5 \mathrm{e}^{-\mathrm{x}}

, where

\mathrm{x}

is the quartity produced. Determin: the total cost producing

5

motor cars if the fixed cost is Rs.

7

lakhs. It is given that

\mathrm{e}^{-5^{2}}=0.006

.

Given Cost Function: The additional cost of producing $x$ motor cars is given by the function $6 + 4x^2 + 1.5e^{-x}$ . We need to calculate this for $x = 5$ .
Calculate Variable Cost: First, we calculate the variable cost for producing $5$ motor cars using the given function: $6 + 4(5)^2 + 1.5e^{(-5)}$ .
Substitute Values: Plugging in the values, we get: $6 + 4(25) + 1.5e^{-5}$ . This simplifies to $6 + 100 + 1.5e^{-5}$ .
Calculate $e^{(-5^2)}$ : We are given that $e^{(-5^2)} = e^{-25} = 0.006$ . We can use this value to calculate $1.5e^{(-5)}$ .
Calculate $1.5e^{-5}$ : Since $e^{-5^2}$ is not the same as $e^{-5}$ , we cannot use the given value of $0.006$ directly. We need to calculate $1.5e^{-5}$ separately.

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