Hello!
Suppose that the price per unit in dollars of a cell phone production is modeled by
p = $55 − 0.0125x,
where x is in thousands of phones produced, and the revenue represented by thousands of dollars is R = x · p.
Find the production level that will maximize revenue.
Well, calculating the production level that maximizes revenue is no laughing matter, but I'll give it a shot!
To find the production level that maximizes revenue, we need to determine the value of x that corresponds to the maximum value of R.
First, let's express R in terms of x:
R = x · p
= x · ($55 − 0.0125x)
Next, we can expand and rearrange the equation:
R = $55x − 0.0125x²
Since we're looking for the maximum value of R, we can differentiate R with respect to x:
dR/dx = 55 − 0.025x
To find the production level that maximizes revenue, we set dR/dx equal to zero and solve for x:
55 - 0.025x = 0
0.025x = 55
x ≈ 2200
So, the production level that will maximize revenue is approximately 2200 thousand phones.
Just remember, this is all just a calculated guess – it's not like the phones will suddenly become hilarious once you hit that production level!
To find the production level that will maximize revenue, we can use the formula for revenue, R = x · p, and manipulate it to maximize the expression.
First, let's substitute the expression for "p" into the formula for revenue:
R = x · (55 - 0.0125x)
Now, let's expand and simplify the expression:
R = 55x - 0.0125x^2
To maximize revenue, we need to find the value of x that will give us the maximum value of R. We can do this by finding the vertex of the equation.
The vertex of a quadratic equation in the form ax^2 + bx + c can be found using the formula:
x = -b / (2a)
In our case, a = -0.0125 and b = 55, so let's substitute these values into the formula:
x = -55 / (2 * -0.0125)
Simplifying:
x = -55 / -0.025
x = 2200
So, the production level that will maximize revenue is 2200 thousands of phones produced.
To find the production level that will maximize revenue, we need to first determine the revenue function and then differentiate it with respect to x. Let's go step by step:
1. Recall that the revenue function is given by R = x · p, where R represents revenue in thousands of dollars and x represents the production level in thousands of phones produced.
2. Substitute the expression for p into the revenue function:
R = x · (55 - 0.0125x)
3. Expand the expression:
R = 55x - 0.0125x²
4. Now, differentiate the revenue function with respect to x to find its critical points.
dR/dx = 55 - 0.025x
5. Set the derivative equal to zero to find the critical points:
55 - 0.025x = 0
6. Solve for x:
0.025x = 55
x = 55 / 0.025
x = 2200
7. The production level that maximizes revenue is obtained at x = 2200.
Therefore, producing 2200 thousand phones will result in maximum revenue.
According to the condition, the revenue (measured in thousands of dollars) is
R(x) = x*(35-0.0125x),
where x is the number of the phones measured in thousand of units.
So, you need to find the maximum of this quadratic function
R(x) = -0.0125x^2 + 35x.
The maximum is achieved at x = -b%2F%282a%29 ( referring to the general form of a quadratic function q(x) = ax%5E2+%2B+bx+%2B+c ),
which at given conditions is x = -35%2F%282%2A%28-0.0125%29%29 = 35%2F0.025 = 1400.
So, the maximum is achieved at the production level 1400 thousand of phone units .
The maximum revenue is the value R(x) at this value of x:
R%5Bmax%5D = R(1400) = -0.0125%2A1400%5E2+%2B+35%2A1400 = 24500 thousands of dollars.
Answer. The maximum revenue is 24500 thousands of dollars achieved at the production level of 1400 thousand of phone units .
Hope this help you with your homework :3