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

From the question, we have:

$Plates = 300$

$Price = 20$

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

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

The slope (m) is calculated using:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

So, we have:

$m = \frac{25-20}{290-300}$

$m = \frac{5}{-10}$

$m = -\frac{1}{2}$

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:

$R(x) = (300 - 10x) \times (20 + 5x)$

Open the brackets to calculate the maximum revenue

$R(x) =6000 + 1500x - 200x - 50x^2$

$R(x) =6000 + 1300x - 50x^2$

Equate to 0

$6000 + 1300x - 50x^2 =0$

Differentiate with respect to x

$1300 - 100x =0$

Collect like terms

$100x =1300$

Divide by 100

$x =13$

So, the price at maximum revenue is:

$Price= 20 + 5x$

$Price= 20 + 5 * 13$

$Price= 85$

In conclusion:

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