Suppose the revenue (in thousands of dollars) for the sales of 'x' hundred units of an electronic item is given by the function R(x)= 40x^2e^ -0.4x +30. where the maximum capacity of the plant is 800 units. Determine the number of units to produce in order to maximize revenue.
I read you revenue function as
R(x)= 40x2 *e-0.4x +30
dR/dx=40*x^2 *(-.4)e^-.4x+80x*e^(-.4x) =0
or x^2 *( .4)e^-.4x =2 x*e^(-.4x)
x( .4)=2
x=5 hundred units.
check my work.
R(x)= 40x^2 { e^ ???? } -0.4x +30
Well, let's start by finding the derivative of the revenue function to locate its critical points. Then we can figure out where the revenue is maximized.
The derivative of R(x) with respect to x is dR/dx = 80xe^(-0.4x) - 16x^2e^(-0.4x), obtained using the product and chain rule.
To find the critical points, we need to solve for when the derivative equals zero:
0 = 80xe^(-0.4x) - 16x^2e^(-0.4x)
Dividing through by 16x^2e^(-0.4x), we get:
0 = 5 - x
Solving for x, we find x = 5. So, one critical point is x = 5.
Now let's check the endpoints of our possible range. Since the maximum capacity of the plant is 800 units, the revenue function is only valid for x ≤ 8 (as we're assuming x is in hundreds).
When x = 0, the revenue is R(0) = 40(0)^2e^(-0.4(0)) + 30 = 30.
When x = 8, the revenue is R(8) = 40(8)^2e^(-0.4(8)) + 30 = 2285.99.
To determine the maximum revenue, we just need to compare the revenue at x = 5 (our critical point), and the endpoints x = 0 and x = 8.
We have R(0) = 30, R(5) ≈ 1348.47, and R(8) ≈ 2285.99.
Since R(8) is the highest revenue among those three values, the maximum revenue is achieved when approximately 800 units (x = 8) are produced.
So, to maximize revenue, you should produce 800 units. Just make sure to have enough clowns on hand to handle the production!
To determine the number of units to produce in order to maximize revenue, we need to find the maximum value of the revenue function.
Given: R(x) = 40x^2 * e^(-0.4x) + 30
To find the maximum value, let's differentiate the revenue function with respect to x.
dR/dx = 80x * e^(-0.4x) + 40x^2 * (-0.4) * e^(-0.4x)
Next, set the derivative equal to zero to find the critical points:
0 = 80x * e^(-0.4x) + 40x^2 * (-0.4) * e^(-0.4x)
Simplifying the equation:
0 = 80x * e^(-0.4x) - 16x^2 * e^(-0.4x)
Now, we can factor out e^(-0.4x):
0 = e^(-0.4x) * (80x - 16x^2)
Setting each factor equal to zero:
e^(-0.4x) = 0 (No real solutions)
80x - 16x^2 = 0
Factoring out 16x:
16x(5 - x) = 0
Setting each factor equal to zero:
16x = 0
x = 0 (This is not valid since we can't produce zero units.)
5 - x = 0
x = 5
From the equation, we have two critical points: x = 0 and x = 5.
However, since the maximum capacity of the plant is 800 units (800/100 = 8), we need to choose a value between 0 and 5.
Next, we can use the second derivative test to determine if x = 5 gives us a maximum or minimum.
Taking the second derivative of the revenue function:
d^2R/dx^2 = 80 * e^(-0.4x) - 16x * e^(-0.4x) * (-0.4) - 16x * e^(-0.4x) * (-0.4) + 40x^2 * (-0.4) * (-0.4) * e^(-0.4x)
Simplify:
d^2R/dx^2 = 80 * e^(-0.4x) + 6.4x * e^(-0.4x) + 2.56x^2 * e^(-0.4x)
Now, substitute x = 5 into the second derivative:
d^2R/dx^2 = 80 * e^(-0.4 * 5) + 6.4(5) * e^(-0.4 * 5) + 2.56(5^2) * e^(-0.4 * 5)
Evaluate the expression:
d^2R/dx^2 ≈ 80 * e^-2 + 6.4(5) * e^-2 + 2.56(5^2) * e^-2
≈ 80 * 0.135 + 6.4(5) * 0.135 + 2.56(5^2) * 0.135
≈ 10.8 + 10.8 + 34.2
≈ 55.8
Since the second derivative is positive (55.8 > 0), this confirms that x = 5 is a local minimum.
Therefore, the number of units to produce in order to maximize revenue is x = 5 (500 units).
To determine the number of units to produce in order to maximize revenue, you need to find the value of 'x' that results in the maximum value for the revenue function R(x).
Step 1: Determine the domain of the function
The domain of the revenue function R(x) is the set of possible input values for 'x'. In this case, the maximum capacity of the plant is 800 units. Therefore, the domain of the function is x ≥ 0 and x ≤ 800.
Step 2: Find the critical points
To find the critical points of the function R(x), you need to find the values of 'x' for which the derivative of the function is either zero or undefined. Start by calculating the derivative of R(x) using the product rule and the chain rule:
R'(x) = (80x - 8x^2)e^(-0.4x)
Step 3: Set the derivative equal to zero and solve for 'x'
To find the critical points, set R'(x) = 0 and solve for 'x':
(80x - 8x^2)e^(-0.4x) = 0
Since the exponential term e^(-0.4x) is always positive, the equation can be simplified to:
80x - 8x^2 = 0
Divide the equation by 8x:
10 - x = 0
Solving for 'x' gives x = 10.
Step 4: Check the endpoints of the domain
In this case, the domain is x ≥ 0 and x ≤ 800. Since x = 10 lies within the domain, there is no need to check the endpoints.
Step 5: Evaluate the function at the critical points and endpoints
To determine whether the critical point or endpoints yield the maximum revenue, evaluate the revenue function at these values:
R(10) = 40(10^2)e^(-0.4(10)) + 30
Calculating the expression would yield the maximum revenue at these production levels.
Therefore, to maximize revenue, the number of units to produce would be 10 hundred units.