word problem

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xtyna

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Oct 20, 2010
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help!
Tshirts sell for $50. Usually 400 are sold in 1 week.
if the price increases by $10, then 5 less are sold.
a. what is the usual weekly Revenue?
b. what is the equation, define the variables, convert to standard form.
c. what is the max, show using algebra

my work....
c=, s= sales in 1 week, R= revenue

"Usual" means AVerage?????
therefore average weekly sales = cs + (c+10)(s-5)
2
Not really sure what my unknowns are
 


I would ignore the adjective "usual".

I mean, the usual sales volume is 400 shirts per week, but that number changes with each price hike.

The unknown in this scenario is how many $10 price increases there are this week.

The shirt price and the number sold are both functions of the number of price hikes; therefore, so is the revenue.

They want you to write an expression for an arbitrary week's revenue, in terms of the number of price hikes for that week.

Let n = the number of $10-price-hikes applied to the $50 base price

Now we can write the following two expressions.

The shirt price during an arbitrary week with n price hikes is 50 + 10n.

Does that makes sense ? $10 gets added to the $50, for each price hike.

The number of shirts sold during the week is 400 - 5n.

Each hike causes five sales to be subtracted from the week's usual 400 sales.

Okay, I suppose this tells us that the adjective "usual" means "when there are no price hikes". In other words, it's "usual" for n to be 0, and, hence, it's usual for the shirt price to be $50 and the week's sales to be 400.

The equation for revenue is:

Revenue = Number of sales × Price per sale

With this information, can you continue ?


My EDIT: Added comment that "usual" means n = 0
 
\(\displaystyle Week: \ \ Price \ \ X \ \ number \ sold \ = \ Total \ Value.\)

\(\displaystyle 1 \ \ \ \ \ [\$50+(10)(0) ] \ X \ [400-(5)(0)] \ = \ \$20.000.\)

\(\displaystyle 2 \ \ \ \ \ [\$50+(10)(1)] \ X \ [400-(5)(1)] \ = \ \$23,700.\)

\(\displaystyle 3 \ \ \ \ \ [\$50+(10)(2)] \ \ \ X \ [400-(5)(2)] \ = \ \$27,300.\)

\(\displaystyle n \ \ \ \ \ [\$50+(10)(n-1) \ \ \ X \ [400-(5)(n-1)] \ = \ -50n^2+3850n+16,200.\)

\(\displaystyle f(n) \ = \ -50n^2+3850n+16,200, \ n \ = \ week.\)

\(\displaystyle f'(n) \ = \ -100n+3850 \ = \ 0, \ \implies \ n \ = \ 38.5, \ rounded \ to \ 38th \ week, \ 0 \ < \ n \ < \ 81, \ n \ an \ integer.\)

\(\displaystyle Check: \ f(37) \ = \ \$90,200.\)

\(\displaystyle f(38) \ = \ \$90,300.\)

\(\displaystyle f(40) \ = \ \$90,200.\)
 


Glenn took a somewhat different approach, but our end-results are the same.

I note that 38 ten-dollar hikes in the t-shirt price means that their sales drop to nearly half the usual number, and they charge $430 for each t-shirt.

 
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