Answer:
Let x = number of $1 rise in taco price
TP = Taco Price in dollars
A = Average number of Tacos Sold
Complete table will be
x | TP | A | Daily Revenue ($)
0 | 4 | 60 | 240
1 | 5 | 53 | 265
2 | 6 | 46 | 276
3 | 7 | 39 | 273
4 | 8 | 32 | 256
5 | 9 | 25 | 225
- Optimal Taco price = $6
Explanation:
Complete Question
Tico’s taco truck is trying to determine the best price at which to sell tacos, the only item on the menu, to maximize profits. The taco trucks owner decided to adjust the price per taco and record data on the number of tacos sold each day with each new price. When the taco truck charges $4 for a taco, it sells an average of 60 tacos in one day. With every $1 increase in the price of a taco, the number of tacos sold per day decreases by 7.
The owner can calculate the daily revenue using the polynomial expression (-7x²+32x+240),
where x is the number of $1 increases in the taco price. In this activity, you’ll interpret and manipulate this expression and the scenario it represents.
Complete the table for different values of x in the polynomial expression (-7x²+32x+240) then, determine the optimal price that the taco truck should sell it’s tacos for. Assume whole dollar amounts for the tacos.
Solution
The original price of taco = $4
x = number of $1 price increase of taco
Taco price = $(4+x)
Original demand of taco = 60
For each $1 rise in taco price, number of tacos sold per day falls by 7 units
Average number of tacos sold = (60 - 7x)
Revenue made from selling taco at a price increase of x is given as
R(x) = (-7x²+32x+240)
Let x = number of $1 rise in taco price
TP = Taco Price in dollars
A = Average number of Tacos Sold
x | TP | A | Daily Revenue ($)
0 | 4 | 60 | 240
1 | 5 | 53 | 265
2 | 6 | 46 | 276
3 | 7 | 39 | 273
4 | 8 | 32 | 256
5 | 9 | 25 | 225
From the table, the optimal price, which evidently maximizes the revenue, is $6.
But, we will still use differentiation analysis to solve this.
At the optimal price, the revenue function will be at the highest. At the highest point,
(dR/dx) = 0 and (d²R/dx²) < 0
R(x) = (-7x²+32x+240)
(dR/dx) = -14x + 32 = 0
x = (32/14) = 2.286
(d²R/dx²) = -14 < 0
Hence, the maximum for the revenue function occurs when the number of $1 increase in price of taco is 2.286.
Optimal price of taco is the $(4+2.286) = $6.286 = $6 to the nearest whole number.
This is the same as the value obtained from the table.
Hope this Helps!!!