Get reliable answers to your questions at Westonci.ca, where our knowledgeable community is always ready to help. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

A restaurant operator in Accra has found out that during the lockdown,if she sells a plate of her food for Ghc 20 each,she can sell 300 plates but for each Ghc5 she raises the price,10 less plates are sold. A.Draw a table relating 5 different price levels with their corresponding number of plates sold.

B.Use the table to find the slope of the demand

C.Find the equation of the demand fraction.

D.Use your equation to determine the price in Ghc if she sells one plate of food to maximize her revenue

Sagot :

The demand equation illustrates the price of an item and how it relates to the demand of the item.

  • The slope of the demand function is -1/2
  • The equation of the demand function is: [tex]R(x) = (300 - 10x) \times (20 + 5x)[/tex]
  • The price that maximizes her revenue is: Ghc 85

From the question, we have:

[tex]Plates = 300[/tex]

[tex]Price = 20[/tex]

The number of plates (x) decreases by 10, while the price (y) increases by 5. The table of value is:

[tex]\begin{array}{cccccc}x & {300} & {290} & {280} & {270} & {260} \ \\ y & {20} & {25} & {30} & {35} & {40} \ \end{array}[/tex]

The slope (m) is calculated using:

[tex]m = \frac{y_2 - y_1}{x_2 - x_1}[/tex]

So, we have:

[tex]m = \frac{25-20}{290-300}[/tex]

[tex]m = \frac{5}{-10}[/tex]

[tex]m = -\frac{1}{2}[/tex]

The equation of the demand is as follows:

The initial number of plates (300) decreases by 10 is represented as: (300 - 10x).

Similarly, the initial price (20) increases by 5 is represented as: (20 + 5x).

So, the demand equation is:

[tex]R(x) = (300 - 10x) \times (20 + 5x)[/tex]

Open the brackets to calculate the maximum revenue

[tex]R(x) =6000 + 1500x - 200x - 50x^2[/tex]

[tex]R(x) =6000 + 1300x - 50x^2[/tex]

Equate to 0

[tex]6000 + 1300x - 50x^2 =0[/tex]

Differentiate with respect to x

[tex]1300 - 100x =0[/tex]

Collect like terms

[tex]100x =1300[/tex]

Divide by 100

[tex]x =13[/tex]

So, the price at maximum revenue is:

[tex]Price= 20 + 5x[/tex]

[tex]Price= 20 + 5 * 13[/tex]

[tex]Price= 85[/tex]

In conclusion:

  • The slope of the demand function is -1/2
  • The equation of the demand function is: [tex]R(x) = (300 - 10x) \times (20 + 5x)[/tex]
  • The price that maximizes her revenue is: Ghc 85

Read more about demand equations at:

https://brainly.com/question/21586143

Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.