The profit function of a company can be represented by p(x) = x –0.00001x2, where x is unit
sold. Find the optimal sales volume and the amount of profit to be expected at that volume.
max p(x) occurs at the vertex of the parabola, which is at
x = -b/2a = 1/0.00002 = 50,000
Ah, the wonderful world of quadratic profit functions! Let me put on my clown nose and crunch some numbers for you.
To find the optimal sales volume, we need to find the maximum point on the profit function. The maximum point occurs at the vertex of the parabola formed by the profit function. The formula for the x-coordinate of the vertex of a quadratic function in the form ax^2 + bx + c is given by x = -b/(2a).
In your case, the profit function p(x) = x - 0.00001x^2 is already in the form ax^2 + bx + c, with a = -0.00001, b = 1, and c = 0. Plugging these values into the formula, we get x = -1/(2*-0.00001) = 50,000.
So, the optimal sales volume for maximum profit is 50,000 units. Now, let's calculate the amount of profit to be expected at that volume.
Plugging x = 50,000 into the profit function p(x), we get:
p(50,000) = 50,000 - 0.00001(50,000)^2
= 50,000 - 0.00001(2,500,000,000)
= 50,000 - 25,000
= 25,000
So, at the optimal sales volume of 50,000 units, the expected profit is $25,000.
But remember, these calculations are purely based on the mathematical model. In the real world, many other factors come into play, like costs, competition, and market demand. So, take these results with a pinch of clown sprinkles!
To find the optimal sales volume and the amount of profit at that volume, we need to maximize the profit function p(x) = x - 0.00001x^2.
Let's start by taking the derivative of the profit function with respect to x:
p'(x) = 1 - 0.00002x
To find the critical points, we set p'(x) equal to zero and solve for x:
1 - 0.00002x = 0
0.00002x = 1
x = 1 / 0.00002
x ≈ 50,000
Now that we have the critical point, we need to determine whether it is a maximum or minimum. To do this, we can take the second derivative of the profit function:
p''(x) = -0.00002
Since p''(x) is negative for all values of x, we can conclude that x ≈ 50,000 is a maximum point.
So, the optimal sales volume is approximately 50,000 units. To find the amount of profit at that volume, we substitute x = 50,000 into the profit function:
p(50,000) = 50,000 - 0.00001(50,000)^2
p(50,000) = 50,000 - 0.00001(2,500,000,000)
p(50,000) = 50,000 - 25,000
p(50,000) = 25,000
Therefore, the amount of profit expected at the optimal sales volume of 50,000 units is $25,000.
To find the optimal sales volume and the amount of profit to be expected at that volume, we need to maximize the profit function p(x) = x - 0.00001x^2.
To do this, we can find the derivative of the profit function with respect to x and set it equal to zero, and then solve for x.
First, let's find the derivative of the profit function p(x):
p'(x) = 1 - 2 * 0.00001x
Next, set p'(x) equal to zero:
1 - 2 * 0.00001x = 0
Solving this equation for x will give us the optimal sales volume.
1 - 2 * 0.00001x = 0
1 = 2 * 0.00001x
x = 1 / (2 * 0.00001)
x ≈ 50,000
So, the optimal sales volume is approximately 50,000 units.
To find the amount of profit expected at this volume, we substitute this value of x into the profit function p(x):
p(50,000) = 50,000 - 0.00001 * (50,000)^2
p(50,000) ≈ 50,000 - 2500
p(50,000) ≈ 47,500
Therefore, at the optimal sales volume of 50,000 units, the expected profit is approximately $47,500.