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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
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