Hey! Im stuck on my uni calculus assignment... couple of questions i just cant get my head round! if anyone can help id be forever grateful :-)
11. A shop owner needs to calculate the cost of an item including the sales tax of 12%.
Before tax, the shop gives a discount of 8% on the original list price of the product.
(Total 7 marks; 2,3,2)
a. Write an equation that connects the total cost C of the product with the list price X.
b. If the shop owner initially puts a mark up of 100% to arrive at the list price,
what is the net profit or loss for the owner. Assume a list price of $100.
c. Using the symbols above, write the equation for net profit.
16. A certain medicine reduces the bacteria present in a person by 10% every hour.
Initially 100,000 bacteria are present. Make a table of values for the number of bacteria
present at the end of each hour for 8 hours.
a. Plot the graph of Time Vs Bacterial count on a graph paper. Use appropriate scales.
b. Approximately in how many hours will the bacterial count drop to 25000?
c. Use algebraic method to calculate the time (in hours) when the bacterial count will be 1000 or below?
"11. A shop owner needs to calculate the cost of an item including the sales tax of 12%. Before tax, the shop gives a discount of 8% on the original list price of the product.
a. Write an equation that connects the total cost C of the product with the list price X.
b. If the shop owner initially puts a mark up of 100% to arrive at the list price,
what is the net profit or loss for the owner. Assume a list price of $100.
c. Using the symbols above, write the equation for net profit."
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11a. C = (1.12)(X - 0.08X)
Therefore, C = 1.0304X
11b. NP = (1 - 0.08)($100) - (0.50)($100) = $42
11c. NP = (1 - 0.08)X - (0.50)X = 0.42X
Sales tax is not a factor in Items 11b and 11c because the shop owner must pass this money on to the government.
The bacterial problem is one of those simple differential equations that can be solved by inspection!
The change in the number of bacteria is porportional to the number present and a certain t.
Y' = -kY.
thus Y = exp(-kt) (minus because numbers are dropping, k is positive)
For your problem, Y' = -0.1Y (the change in the number of bacteria is less 10% of the current amount!)
Thus Y = Y0exp(-0.1t)
The Y0 because at t=0, exp(-0.1t)=1
What is the value at t=0? 100,000 as stated in the question.
Thus the equation for the number of bacteria with respect to time is
Y = 100,000 exp (-0.1t)
To find the number of bacteria at any time you choose, it suffices to plug in your chosen value of t. If you want to find out at what time there is a certain amount of bacteria, replace the number by Y and solve the equation for t! Easy!
For the actual question:
1. What time does the number drop to 2,500:
25,000 = 100,000 exp (-0.1t)
2500/100000 = exp (-0.1t)
ln (25,000/100,000) = -0.1t
(ln (25,000/100,000))/-0.1 = t
t = 13.8629 hours
2. What time does the number drop to 1,000:
(ln (1,000/100,000))/-0.1 = t (using the working from previous question)
t = 46.0517 hours
Would you like to see a graph of the number of bacteria versus time? I made one in Graphmatica:
Last edited by Quantumcat; Jun 18th, 2007 at 10:14 PM.
Calculus!!!
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