Draw
Question 1
It is estimated that [tex]$t$[/tex] years from now, the circulation of a local newspaper will be
[tex]$C(t)=100 t^2+400 t+5000$[/tex]
(a) Derive an expression for the rate at which the circulation will be changing with respect to time [tex]$t$[/tex] years from now.
(b) At what rate will the circulation be changing with respect to time 5 years from now? Will the circulation be increasing or decreasing at that time?
(c) By how much will the circulation actually change during the slith years?
Question 2
A company manufactures and sells [tex]$x$[/tex] transistor radios per week. If the weekly cost and price - demand equations are
[tex]
\begin{array}{l}
C(x)=5000+2 x \\
P=10-\frac{x}{1000}, 0 \leq x \leq 8000
\end{array}
[/tex]
The government has decided to tax the company Ghe 2.00 for each radio produced. Taking into consideration this additional cost.
(i) How many radios should the company manufacture in order to maximize its weekly profit?
(ii) What is the maximum weekly profit?
(iii) How much should it charge for the radios?
Question 3
A manufacturer of motorcycles estimates that if [tex]$x$[/tex] thousand Ghana cedi is spent on advertising, then [tex]$M(x)=2300+\frac{125}{x}-\frac{517}{x^2}$[/tex], for [tex]$3 \leq x \leq 18$[/tex]
Cycles will be sold.
(a) At what rate will sales be changing when Ghe 9,000 is spent on advertising? Are sales increasing or decreasing for this level of advertising expenditure?
Answer (2)
Find the derivative of the circulation function: C ′ ( t ) = 200 t + 400 .
Evaluate the derivative at t = 5 : C ′ ( 5 ) = 1400 , indicating an increasing circulation.
Calculate the change in circulation during the sixth year: C ( 6 ) − C ( 5 ) = 1500 .
Determine the optimal production level for maximum profit: x = 3000 , with a maximum profit of Ghe 4000 and a price of Ghe 7 per radio. Sales are decreasing at a rate of -0.1248 when Ghe 9,000 is spent on advertising.
Explanation
Understanding the Problem We are given the circulation function C ( t ) = 100 t 2 + 400 t + 5000 . We need to find the rate of change of circulation with respect to time, the rate of change at t = 5 , and the actual change in circulation during the sixth year.
Finding the Derivative To find the rate of change of circulation, we need to find the derivative of C ( t ) with respect to t . Using the power rule, we have C ′ ( t ) = d t d ( 100 t 2 + 400 t + 5000 ) = 200 t + 400
Rate of Change at t=5 Now, we need to find the rate of change at t = 5 . We substitute t = 5 into C ′ ( t ) :
C ′ ( 5 ) = 200 ( 5 ) + 400 = 1000 + 400 = 1400 Since 0"> C ′ ( 5 ) = 1400 > 0 , the circulation is increasing at t = 5 .
Change in Circulation During the Sixth Year To find the actual change in circulation during the sixth year, we need to calculate C ( 6 ) − C ( 5 ) .
First, we find C ( 6 ) :
C ( 6 ) = 100 ( 6 ) 2 + 400 ( 6 ) + 5000 = 100 ( 36 ) + 2400 + 5000 = 3600 + 2400 + 5000 = 11000 Next, we find C ( 5 ) :
C ( 5 ) = 100 ( 5 ) 2 + 400 ( 5 ) + 5000 = 100 ( 25 ) + 2000 + 5000 = 2500 + 2000 + 5000 = 9500 Then, we calculate the difference: C ( 6 ) − C ( 5 ) = 11000 − 9500 = 1500
Setting up Profit Maximization For Question 2, we have the cost function C ( x ) = 5000 + 2 x and the price function P = 10 − 1000 x . The government imposes a tax of Ghe 2.00 per radio, so the new cost function is C t ( x ) = 5000 + 2 x + 2 x = 5000 + 4 x .
The revenue function is R ( x ) = x P = x ( 10 − 1000 x ) = 10 x − 1000 x 2 .
The profit function is P ( x ) = R ( x ) − C t ( x ) = ( 10 x − 1000 x 2 ) − ( 5000 + 4 x ) = 6 x − 1000 x 2 − 5000 .
Finding Critical Points To maximize the profit, we find the derivative of P ( x ) and set it to zero: P ′ ( x ) = d x d ( 6 x − 1000 x 2 − 5000 ) = 6 − 1000 2 x = 6 − 500 x Setting P ′ ( x ) = 0 , we get 6 − 500 x = 0 , which gives x = 3000 .
Calculating Maximum Profit and Price To ensure that x = 3000 maximizes the profit, we check the second derivative: P ′′ ( x ) = d x d ( 6 − 500 x ) = − 500 1 Since P ′′ ( x ) < 0 , the profit is maximized at x = 3000 .
The maximum weekly profit is: P ( 3000 ) = 6 ( 3000 ) − 1000 ( 3000 ) 2 − 5000 = 18000 − 9000 − 5000 = 4000 The price per radio is: P = 10 − 1000 3000 = 10 − 3 = 7
Analyzing Sales Rate For Question 3, we have the sales function M ( x ) = 2300 + x 125 − x 2 517 . We need to find the rate of change of sales when Ghe 9,000 is spent on advertising, which means x = 9 .
First, we find the derivative of M ( x ) with respect to x :
M ′ ( x ) = d x d ( 2300 + x 125 − x 2 517 ) = − x 2 125 + x 3 2 ( 517 ) = − x 2 125 + x 3 1034 Now, we substitute x = 9 into M ′ ( x ) :
M ′ ( 9 ) = − 9 2 125 + 9 3 1034 = − 81 125 + 729 1034 = − 1.5432 + 1.4184 = − 0.1248 Since M ′ ( 9 ) < 0 , the sales are decreasing when Ghe 9,000 is spent on advertising.
Final Answers Question 1: (a) The rate at which the circulation will be changing is C ′ ( t ) = 200 t + 400 .
(b) At t = 5 , the rate is 1400, and the circulation is increasing. (c) The circulation will actually change by 1500 during the sixth year. Question 2: (i) The company should manufacture 3000 radios to maximize its weekly profit. (ii) The maximum weekly profit is Ghe 4000. (iii) It should charge Ghe 7 for the radios. Question 3: (a) At Ghe 9,000 spent on advertising, the sales are changing at a rate of -0.1248, and sales are decreasing.
Examples
Understanding rates of change is crucial in business. For example, knowing how the circulation of a newspaper changes over time helps in predicting future readership and planning advertising strategies. Similarly, determining the optimal production level to maximize profit, considering factors like cost, price, and taxes, is essential for business success. Analyzing the impact of advertising expenditure on sales enables businesses to make informed decisions about their marketing budgets.
A current of 15.0 A for 30 seconds corresponds to a total of 450 coulombs of charge. This charge is equivalent to approximately 2.81 billion billion electrons flow through the device. The calculation involves using the relationship between current, charge, and the charge of a single electron.
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