When priced at 300 pesos each, a toy has an annual sales of 4000 units. The manufacturer estimates that each 10 pesos increase per unit will decrease the sales by 100 units. Find the unit price that will maximize the total revenue.

sales=4000-100((p-300)/10 where p is the price.

sales=4000-10p+3000

revenue=sales*p= p(7000-10p) so revenue has zeroes at p=0 and p=700, so a max occurs at p=350

check all my math, I did it in my head.

To find the unit price that will maximize the total revenue, we need to determine the relationship between the unit price and the number of units sold. Let's break down the problem step by step:

1. Let's start by defining some variables:
- Let x be the number of 10-peso increases per unit.
- The unit price at the beginning is 300 pesos, so the initial price will be 300 + (10 * x) pesos.
- The number of units sold will be 4000 - (100 * x).

2. The total revenue is calculated by multiplying the unit price by the number of units sold. So, the revenue equation will be:
Revenue = (300 + 10x) * (4000 - 100x)

3. To find the unit price that maximizes the total revenue, we need to find the value of x that maximizes the revenue.

4. We can expand the revenue equation to simplify it further:
Revenue = (300 + 10x) * (4000 - 100x)
= (300 * 4000) + (300 * (-100x)) + (10x * 4000) + (10x * (-100x))
= 1,200,000 + (-30,000x) + 40,000x + (-1,000x^2)
= -1,000x^2 + 10,000x + 1,200,000

5. To maximize the revenue, we need to find the maximum point on the graph of the revenue equation. Since the equation is quadratic, we can use the vertex formula to find the x-coordinate of the vertex. The x-coordinate of the vertex can be found using the equation:
x = -b / (2a)

In our case:
a = -1,000
b = 10,000

x = -10,000 / (2 * -1,000)
= -10,000 / -2,000
= 5

6. The maximum revenue occurs when x = 5. We can substitute this value back into the revenue equation to find the maximum revenue:
Revenue = -1,000(5^2) + 10,000(5) + 1,200,000
= -1,000(25) + 50,000 + 1,200,000
= -25,000 + 50,000 + 1,200,000
= 1,225,000

7. Therefore, the unit price that will maximize the total revenue is 300 + (10 * 5) = 350 pesos, and the maximum total revenue will be 1,225,000 pesos.