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insens350 [35]
3 years ago
10

Demand The demand function for a product is given by

Mathematics
1 answer:
stira [4]3 years ago
4 0

Answer:

x= -\frac{ln [\frac{5P}{8000-P}]}{0.002}

a) x= -\frac{ln [\frac{5*200}{8000-200}]}{0.002} =1027.062 \approx 1027

b) x= -\frac{ln [\frac{5*800}{8000-800}]}{0.002} =293.893 \approx 294

Step-by-step explanation:

For this case we have the following function:

P= 8000 (1- \frac{5}{5 +e^{-0.002 x}})

We can solve for x like this. First we can reorder the expression like this:

\frac{P}{8000} = 1- \frac{5}{5+e^{-0.002x}}

\frac{5}{5+e^{-0.002x}} = 1 -\frac{P}{8000} = \frac{8000-P}{8000}

\frac{40000}{8000-P} = 5 + e^{-0.002x}

Now we can apply natura log on both sids and we got:

ln[\frac{40000}{8000-P} -5] = ln e^{-0.002x}

ln [\frac{5P}{8000-P}] = -0.002x

And if we solve for x we got:

x= -\frac{ln [\frac{5P}{8000-P}]}{0.002}

Part a

For this case we can replace P = 200 and see what we got for x like this:

x= -\frac{ln [\frac{5*200}{8000-200}]}{0.002} =1027.062 \approx 1027

Part b

For this case we can replace P = 800 and see what we got for x like this:

x= -\frac{ln [\frac{5*800}{8000-800}]}{0.002} =293.893 \approx 294

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