Question

Yaster Gadgets manufactures and sells x smartphones per week. The weekly price-demand and cost equations are, respectively.

p=507−0.49x and C(x)=20,430+18x
Suppose Yaster Gadgets wants to maximize weekly profit Compute the following quantities.
1. How many phones should be produced each week? _____ phones. Round to 2 decimal places.
2. What price should Jesaki charge for the phones? $ ____ per phone. Round to the nearest cent.
3. What is the maximum weekly profit? $_____per week. Round to the nearest cent.

Answer 1

To maximize weekly profit, Yaster Gadgets should: 1. produce 500 phones, 2. charge $507 per phone, and 3. achieve a maximum weekly profit of approximately $114,920.

To find the quantities that maximize weekly profit, we need to calculate the profit function and then find its critical points. The profit function (P(x)) is given by subtracting the cost function (C(x)) from the revenue function (R(x)):

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

1. Profit Function:

P(x) = (507 - 0.49x)x - (20,430 + 18x)

2. Simplify the Profit Function:

P(x) = 507x - 0.49x^2 - 20,430 - 18x

3. Combine Like Terms:

P(x) = -0.49x^2 + 489x - 20,430

Now, to find the quantities that maximize profit, we need to find the critical points of P(x) by taking the derivative and setting it equal to zero:

P'(x) = -0.98x + 489

Setting P'(x) = 0:

-0.98x + 489 = 0

Solving for x:

`\[ x = (489)/(0.98) \]\[ x \approx 500 \]`

Now, we need to check the second derivative to confirm it is a maximum:

P''(x) = -0.98

Since P''(x) is negative, x = 500 is a maximum.

Now, plug x = 500 into the profit function to find the maximum profit:

`\[ P(500) = -0.49(500)^2 + 489(500) - 20,430 \]\[ P(500) \approx \$114,920 \]`

So, the answers are:

1. 500 phones

2. $507 per phone

3. $114,920 per week