I have two problems that I can't figure out for my test review...
1. Productivity: Because a new employee must learn an assigned task, production will increase with time. Suppose that for the average new employee, the rate of performance is given by dN/dt=1/2?(t+1) where N is the number of unity completed t hours after beginning a new tak. If 2 units are completed afer 3 hours, how many units are completed after 8 hours?
2.Suppose that the marginal cost for a product is MC=60?(x+1) and its fixed cost is $340.00. If the marginal revenue for the product is MR=80x, find the profit or loss from production and sale of a) 3 units and b) 8 units
the question mark in prob. 2 for the Marginal Cost is supposed to be a square root sign...sorry!
I assume the ? in the first is sqrt also.
dN/dt=1/2 sqrt (t+1)
N= INT above= 1/3 (t+1)^3/2 + C
2= 1/3 (4)^3/2 + C or C= 2-8/3 check that.
THen solve for t=8
2)
Marginal cost=dC/dx
Marginal revenue=dR/dx
Profit=INT MR dx - INT MC dx + FC
Profit= INT 80x dx -INT 60sqrt(x+1)dx + FC
To solve the first problem, you are given the rate of performance, dN/dt, as 1/2√(t+1) representing the number of units completed t hours after beginning a new task. To find the number of units completed after 8 hours, you need to integrate the rate of performance function.
To integrate 1/2√(t+1), you can use the power rule of integration. First, rewrite the function as (t+1)^(1/2)/2. Then apply the power rule of integration, which states that the integral of x^n with respect to x is (x^(n+1))/(n+1), where n is not equal to -1.
Integrating (t+1)^(1/2)/2 with respect to t, you get (t+1)^(3/2)/(3/2) + C. Simplifying further, you have 2/3 * (t+1)^(3/2) + C.
Since 2 units are completed after 3 hours, you can plug in t=3 into the integrated function to find the value of C.
2 = 2/3 * (3+1)^(3/2) + C
Now solve for C:
2 = 2/3 * 4^(3/2) + C
2 = 2/3 * 8 + C
2 = 16/3 + C
C = 2 - 16/3
C = 6/3 - 16/3
C = -10/3
Now that you have the value of C, you can find the number of units completed after 8 hours by plugging t=8 into the integrated function:
N = 2/3 * (8+1)^(3/2) + (-10/3)
Simplifying further,
N = 2/3 * (9)^(3/2) - 10/3
N = 2/3 * 27 - 10/3
N = 54/3 - 10/3
N = 44/3
Therefore, after 8 hours, approximately 14.67 units are completed.
For the second problem, you are given the marginal cost function, MC=60√(x+1), and the marginal revenue function, MR=80x. To find the profit or loss from production and sale of a) 3 units and b) 8 units, you need to integrate the marginal cost and marginal revenue functions accordingly.
The profit formula is given by:
Profit = ∫ MR dx - ∫ MC dx + FC
To find the profit from producing and selling 3 units, you need to integrate the marginal revenue and marginal cost functions from 0 to 3 units:
Profit = ∫(0 to 3) 80x dx - ∫(0 to 3) 60√(x+1) dx + 340
To integrate 80x with respect to x, use the power rule of integration:
∫80x dx = 40x^2 + C
To integrate 60√(x+1) with respect to x, use the substitution method. Let u = x+1, then du = dx:
∫60√(x+1) dx = ∫60√u du = 60(2/3)u^(3/2) + C = 40u^(3/2) + C
Therefore, the profit from producing and selling 3 units is:
Profit = ∫(0 to 3) 40x^2 dx - ∫(0 to 3) 40(x+1)^(3/2) dx + 340
Now substitute the limits of integration and evaluate the integrals:
Profit = [(40/3)x^3]0 to 3 - [(40/5)(x+1)^(5/2)]0 to 3 + 340
Profit = (40/3)(3^3) - (40/5)(4^(5/2) - 1) + 340
Profit = 360 - 1280/5 + 340
Profit = 360 - 256 + 340
Profit = 444 - 256
Profit = 188
Therefore, the profit from producing and selling 3 units is $188.
To find the profit from producing and selling 8 units, you would follow the same steps as above but integrate from 0 to 8 units instead of 3 units. Plug in the limits of integration, evaluate the integrals, and find the profit at the end.