Question

Shakespear's Pizza sells 1 comma 100 large Vegi Pizzas per week for ​$16 a pizza. When the owner offers a​ $5 discount, the weekly sales increase to 1 comma 600. ​(a) Assume a linear relation between the weekly sales​ A(x) and the discount x. Find​ A(x). ​(b) Find the value of x that maximizes the weekly revenue. ​(c) Answer parts​ (a) and​ (b) if the price of one pizza is ​$8 and all other data are unchanged.

Answer 1

Answer:

(a) A(x) = 100x + 1100

(b) x = 2.5 USD

(c) A(x) = 100x + 1100 and x = -1.5 USD

Step-by-step explanation:

(a) Since there's a linear relation between weekly sales A(x) and the discount, we can model it as the following expression

A(x) = mx + b

where x is the discount, m is the slope and b is the intersect of the linear equation. We can solve for slope first

$m = \frac{s_2 - s_1}{x_2 - x_1} = \frac{1600 - 1100}{5 - 0} = \frac{500}{5} = 100$

To solve for b, we can just plug in m = 100, x = 0 and A(0) = 1100

1100 = 100*0 + b

b = 1100

Therefore A(x) = 100x + 1100

With discount x, the actual price would then be p - x. Where p is the original undiscounted pricing ($16 in this case)

And with sale A(x), the revenue would then be

$R = (p - x)A(x) = (p - x)*(100x + 1100)$

$R = 100xp + 1100p - 100x^2 - 1100x$

$R = -100x^2 + (100p - 1100)x + 1100p$

To find the maximum value of this, we can take the 1st derivative and set it to 0

R' = -200x + 100p - 1100 = 0

200x = 100p - 1100

$x = \frac{100p - 1100}{200} = 0.5p - 5.5$

when p = $16 then x = 0.5*16 - 5.5 = 8 - 5.5 = 2.5 USD

(c) when the price p is $8 then A(x) would sill be 100x + 1100 because it doesn't depend on p, but the discount. But to maximize the revenue, x will need to be

0.5p - 5.5 = 0.5*8 - 5.5 = 4 - 5.5 = -1.5 USD

So the Pizza owner would need to raise the price by 1.5 USD because originally the pizza is already inexpensive.

Answer 2

a).  Function for the weekly sales → A(x) = 100x + 1100

b).  For the discount of $2.5 revenue will be maximum.

c).  For the maximum revenue, price of the pizza should be increased by  

     $1.5

Maximization of a linear function:

• Let the linear equation representing the relation between the weekly sales A(x) and the discount 'x' is,

       A(x) = mx + b

       Here, m = sales increase per week

       And x = discount offered

• Revenue R(x) = Sale price × Weekly sales

• To maximize the function find the derivative of revenue R(x) and equate it to zero.

        Find the value of 'x' which represents the discount at which

        maximum revenue was generated.

a). If we plot the graph of the linear equation, two points (0, 1100) and

    (5, 1600) will lie on the graph.

    Slope of the line passing through points = $\frac{y_2-y_1}{x_2-x_1}$

                                                                         = $\frac{1600-1100}{5-0}$

                                                                         = 100

    y-intercept of the graph (at discount 'x' = 0) = 1100

    Therefore, function representing the graph will be,

    A(x) = 100x + 1100

b). Function for the revenue will be,

    R(x) = Weekly sales × price after discount

           = A(x) × (16 - x)

           = (100x + 1100)(16 - x)

    R(x) = 100x(16 - x) + 1100(16 - x)

            = 1600x - 100x² + 17600 - 1100x

    R(x) = -100x² + 500x + 17600

    Derivative of the revenue function and equate it to zero,

    R'(x) = -200x + 500

    -200x + 500 = 0

     x = $2.5

    Therefore, for the discount of $2.5, revenue will be maximum.

c). If the price of one pizza = $8

     Function for the revenue → R(x) = (100x + 1100)(8 - x)

     R(x) = 800x + 8800 - 100x² - 1100x

     R(x) = -100x² - 300x + 8800

     R'(x) = -200x - 300

     For R'(x) = 0,

     -200x - 300 = 0

     x = -$(1.5)

     Therefore, for the maximum revenue, price of the pizza should be  

     increased by $1.5

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