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
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
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
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.