Question

p=544-0.4x and the cost is given by C(x)= 19505+18x, how many phones should be produced to maximize revenue

Answer 1

To maximize revenue, approximately 5937 phones should be produced.

The revenue (R) is given by the product of the quantity x and the price per unit p, which is R = px. In this case,
`\(p = 544 - 0.4x\)`.

The cost function C(x) is given by
`\(C(x) = 19505 + 18x\).`

The revenue is maximized when it is equal to the cost, so we can set up the equation:

R = C(x)

Substitute the expressions for R and C(x):

(544 - 0.4x)x = 19505 + 18x

Now, solve for x:

`\[ 544x - 0.4x^2 = 19505 + 18x \]`

Combine like terms:

`\[ 0.4x^2 + 526x - 19505 = 0 \]`

Now, you can solve this quadratic equation for x. You can use the quadratic formula:

`\[ x = (-b \pm √(b^2 - 4ac))/(2a) \]`

In this case, a = 0.4, b = 526, and c = -19505.

`\[ x = (-526 \pm √(526^2 - 4(0.4)(-19505)))/(2(0.4)) \]`

`\[ x = (-526 \pm √(27787636 + 31220))/(0.8) \]`

`\[ x = (-526 \pm √(27818856))/(0.8) \]`

`\[ x = (-526 \pm 5274)/(0.8) \]`

Now, there are two possible solutions:

1.
`\( x_1 = (-526 + 5274)/(0.8) = (4750)/(0.8) = 5937.5 \)`

2.
`\( x_2 = (-526 - 5274)/(0.8) = (-5800)/(0.8) = -7250 \)`

Since the number of phones cannot be negative, the only valid solution is
`\( x = 5937.5 \)`.

Therefore, to maximize revenue, approximately 5937 phones should be produced.

The probable question may be: "How many phones should be produced to maximize revenue given that price per unit p=544-0.4x and cost function C(x)=19505+18x"