Answer to Question #224076 in Calculus for LASHLEY

Question

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

Answer 1

a. Table

Number of Plates

Price per Plate (Ghc.)

300

20

290

25

280

30

270

35

260

40

b. Slope

Gradient = Change X/ Change Y

= (290 – 280)/ (25 - 20)

= 10/5

=2

c. Demand Equation.

First, note this parameter from the question.

We let x = number of $5 increases and number of 10 decreases in plates sold.

Our Demand equation is:

$R(x) = (300-10x) (10+5x)$

d. Maximized Revenue.

$R(x) = 3000 + 1500x - 3000x - 1500x^2$

$R(x) = -1500x^2 - 1500x + 3000 (collect like terms)$

Next we simplify, by dividing through by -1500

$= 1500x^2/1500 - 1500x/1500 + 3000/1500$

$= X^2 - x + 2$

$X^2 - x + 2 = 0$

Next, we find the axis of symmetry using the formula x = -b/(2*a) where b = 1, a = 1

$X = - (-1)/2*1$

$X = 1/2$

Number of GHc 5 increases = $5x1/2 = GHc2.5$

=2.5 + 20 = GHc 22.5 price gives max revenue.