Profit function and differentials

In summary, the first conversation discusses finding the approximate change in profit for a 1-unit change in demand for a company using the profit function P(x)=-390+24x+5x^2-(1/3)x^3. The answer is found using the differential and is approximately -$989976. It is possible for profit to be negative in this case. The second conversation asks for the approximate increase in the area of an oil slick in the shape of a circle when its radius increases from 1.2 miles to 1.4 miles. The answer is 0.48pi miles^2 and is found using linear approximation. When to use differentials and linear approximation depends on the level of accuracy needed, with different
  • #1
rain
11
0
1. The profit function for a compnay is
P(x)=-390+24x+5x^2-(1/3)x^3, where x represents the demand for the product(doghouse). Find the approximate change in profit for a 1-unit change in demand when the demand is at a level of 1000 doghouses. Use the differential.

-my answer that i got is -$989976
am i correct? can profit be negative?

2. An oil slick is in the shape of a circle. Find the approximate increase in the area of the slick if tis radisu increases form 1.2 miles to 1.4miles.

-my answer is 0.48pi miles^2
am i correct?

and one more question, when would you use differentials and when would you use linear approximation?

Thanks for your time.
 
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  • #2
I don't understand the vocabulary of the first, except that negative profit are usually deficits... but for the second, I learned to do it this way : [tex] A(r)=\pi r^2\Rightarrow A(r+\delta)=\pi(r+\delta)^2=\pi(r^2+2r\delta+\delta^2)\approx A(r)+2\pi r\delta[/tex] because [tex]\delta[/tex] is supposed much smaller than r. So that the increase is approx. 2pi*r*delta... which is exactly what you get...If you would have done no approx. you would have get pi*.52...
 
  • #3
o..okay.
so if the question asks for approximation, my answer is correct?If the question just ask for the increase, then .52pi would be correct?
 
  • #4
difference

The Derivative give the 'exact' rate of change, whereas linear approximation may or may not give the 'exact' rate of change.

A differentiable function f(x) may be represented near a point 'a' as a Taylor's series
f(x)= f(a)+ {f'(a)(x-a)+ (f''(a)(x-a)^2)/2!+ (f'''(a)(x-a)^3)/3!+...}
for any x 'near' a.
where ' indicates the order of the derivatives.

The term between {} may be seen as the difference between the value of the function at x and a.

At first order the approximation may be truncated to
f(x)= f(a)+ {f'(a)(x-a)}.

If f(x)=bx+c is a linear function, the approximation is exact since f'(x)=b, and all other derivatives vanish i.e., f''(x)=0, f'''(x)=0...etc. There is no difference between using linear interpolation or differentials in this case.

If f(x) is not a linear function, when you use linear interpolation then you are only approximating the rate of change (with some precision) using a slope of f'(a)= f(x)-f(a)/(x-a). However, if the function f(x) is smooth enough and the difference between x and a is small enough, the approximation may be a good one.

Try to apply this concept to your problem of the circular oil slick and you will see that this is equivalent to kleinwolf explanation and that you're been asked to approximate at first order.
 
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FAQ: Profit function and differentials

What is a profit function?

A profit function is a mathematical equation that describes the relationship between the cost and revenue of a business. It is used to determine the maximum profit that a company can achieve based on different levels of production and sales.

How is a profit function calculated?

A profit function is typically calculated by subtracting the total cost of production from the total revenue. This can be expressed as P(x) = R(x) - C(x), where P is profit, R is revenue, and C is cost.

What is the role of differentials in a profit function?

Differentials are used in a profit function to measure the change in profit as a result of small changes in production and sales. This allows businesses to make more accurate predictions and decisions regarding their operations and profitability.

How can a profit function be optimized?

A profit function can be optimized by finding the critical points, or points where the derivative of the function is equal to zero. These points represent the maximum profit that can be achieved and can be found using methods such as differentiation and optimization techniques.

What are some common applications of profit functions?

Profit functions are commonly used in business and economics to help companies make strategic decisions about production levels, pricing, and resource allocation. They can also be used in financial analysis and forecasting to evaluate the profitability of investments and projects.

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