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snow_lady [41]
4 years ago
5

4. The revenue of a company for a given month is represented as ????(x) = 1,500x − x^2 and its costs as ????(x) = 1,500 + 1,000x

.
What is the selling price, x, of its product that would yield the maximum profit? Show or explain your answer.
Mathematics
1 answer:
mrs_skeptik [129]4 years ago
4 0

Answer:

The profit will be maximum on x = 250.

Step-by-step explanation:

From the given information:

Revenue = 1500x - x²

Cost = 1500 + 1000x

As we know that

Profit = Revenue - Cost ; Let say it equation 1

Then after putting the values of revenue and cost in equation 1 we have:

Profit = (1500x - x²) - (1500 + 1000x)

Profit = 1500x - x² - 1500 - 1000x

Profit = -x² + 500x - 1500

We know that at the max or min the slope of the graph formed by the profit function will be zero, therefore we find the slope of profit function by taking the first derrivative w.r.t. x as under:

d(Profit)/dx = d/dx(-x² + 500x - 1500)

d(Profit)/dx = -2x + 500

By putting the above slope equal to zero we get:

d(Profit)/dx = -2x + 500 = 0

-2x + 500 = 0

-2x = -500

x = 250

Therefore it is concluded that the profit will be maximum when x will be equal to 250.

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t=\frac{345-300}{\frac{200}{\sqrt{100}}}=2.25    

p_v =P(t_{(99)}>2.25)=0.0133  

Conclusion  

If we compare the p value and the significance level assumed \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean is higher than 300 at 1% of significance.  

Step-by-step explanation:

Data given and notation  

\bar X=345 represent the sample mean

s=200 represent the sample standard deviation

n=100 sample size  

\mu_o =300 represent the value that we want to test

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is higher than 300, the system of hypothesis would be:  

Null hypothesis:\mu \leq 300  

Alternative hypothesis:\mu > 300  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{345-300}{\frac{200}{\sqrt{100}}}=2.25    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=100-1=99  

Since is a one right tailed test the p value would be:  

p_v =P(t_{(99)}>2.25)=0.0133  

Conclusion  

If we compare the p value and the significance level assumed \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean is higher than 300 at 1% of significance.  

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