Question

Could you help me to solve the problem below the cost for producing x items is 50x+300 and the revenue for selling x items is 90x−0.5^2 the company makes is how much it takes in (revenue) minus how much it spends (cost). Recall that profit is revenue minus cost. Set up an expression for the profit from producing and selling x items. Find two values of x that will create a profit of $50. Is it possible for the company to make a profit of $2,500? Please explain

Answer 1

Answer:

Profit function: P(x) = -0.5x^2 + 40x - 300

profit of $50: x = 10 and x = 70

NOT possible to make a profit of $2,500, because maximum profit is $500

Step-by-step explanation:

(Assuming the correct revenue function is 90x−0.5x^2)

The cost function is given by:

$C(x) = 50x + 300$

And the revenue function is given by:

$R(x) = 90x - 0.5x^2$

The profit function is given by the revenue minus the cost, so we have:

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

$P(x) = 90x - 0.5x^2 - 50x - 300$

$P(x) = -0.5x^2 + 40x - 300$

To find the points where the profit is $50, we use P(x) = 50 and then find the values of x:

$50 = -0.5x^2 + 40x - 300$

$-0.5x^2 + 40x - 350 = 0$

$x^2 - 80x + 700 = 0$

Using Bhaskara's formula, we have:

$\Delta = b^2 - 4ac = (-80)^2 - 4*700 = 3600$

$x_1 = (-b + \sqrt{\Delta})/2a = (80 + 60)/2 = 70$

$x_2 = (-b - \sqrt{\Delta})/2a = (80 - 60)/2 = 10$

So the values of x that give a profit of $50 are x = 10 and x = 70

To find if it's possible to make a profit of $2,500, we need to find the maximum profit, that is, the maximum of the function P(x).

The maximum value of P(x) is in the vertex. The x-coordinate of the vertex is given by:

$x_v = -b/2a = 80/2 = 40$

Using this value of x, we can find the maximum profit:

$P(40) = -0.5(40)^2 + 40*40 - 300 = 500$

The maximum profit is $500, so it is NOT possible to make a profit of $2,500.