Question

The Pear company sells pPhones. The cost to manufacture a pPhones is

C(x) = -21x² + 45000z+21159 dollars (this includes overhead costs and production costs for each
pPhone). If the company sells ™ pPhones for the maximum price they can fetch, the revenue function will
be R(x) = -30x² + 189000 dollars.
How many pPhones should the Pear company produce and sell to maximimze profit? (Remember that
profit-revenue-cost.)

X=

Answer 1

To find the number of Phones that the Pear company should produce and sell to maximize profit, we need to use the profit function, which is:

P(x) = R(x) - C(x)

where R(x) is the revenue function and C(x) is the cost function.

Substituting the given functions, we get:

P(x) = (-30x² + 189000) - (-21x² + 45000x + 21159)

P(x) = -9x² + 104000x - 21159

To maximize profit, we need to find the value of x that maximizes P(x). To do this, we can take the derivative of P(x) with respect to x and set it equal to zero:

P'(x) = -18x + 104000 = 0

x = 5777.78

Since we can't produce a fractional number of Phones, we should round this down to 5777 Phones.

Therefore, the Pear company should produce and sell 5777 Phones to maximize profit.

Answer 2

Explanation:

To maximize profit, the Pear company needs to find the value of x that maximizes the difference between revenue and cost, which is the profit function P(x) = R(x) - C(x).

P(x) = R(x) - C(x)

P(x) = (-30x² + 189000) - (-21x² + 45000x + 21159)

P(x) = -9x² + 45000x + 157841

To find the value of x that maximizes P(x), we can use the formula x = -b/2a, where a = -9 and b = 45000.

x = -b/2a

x = -45000/(2*(-9))

x = 2500

So, the Pear company should produce and sell 2500 pPhones to maximize profit