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Mathematics 21 Online
OpenStudy (gabylovesyou):

A bakery charges $0.20 for a cookie that is 2 inches in diameter. If the price is proportional to the area, how much do they charge for an extra large cookie that is 6 inches in diameter? Assume that each cookie is shaped of a circle. (MY ANSWER) -----> $1.80 AM I CORRECT??

OpenStudy (gabylovesyou):

or is it 0.60? lol

OpenStudy (anonymous):

I got 60:)

OpenStudy (gabylovesyou):

so its 0.60?

OpenStudy (anonymous):

except that it says its proportional to the area:/ so idk im confused with the wording...

OpenStudy (anonymous):

I think its 60 but hold on a sec

OpenStudy (gabylovesyou):

ok :D

OpenStudy (ash2326):

@Gabylovesyou You're right

OpenStudy (gabylovesyou):

@ash2326 its 1.80?

OpenStudy (ash2326):

Yeah it's $1.80:)

OpenStudy (gabylovesyou):

ok thank you soo much!! :D

OpenStudy (ash2326):

How did you get it?

OpenStudy (apoorvk):

0.60 is incorrect. It's proportional to the *area* not the diameter. so find out the respective area of each type of cookie.

OpenStudy (anonymous):

I was just gonna ask that cuz I tried finding the area and doing it but I still got .6? The area of the first cookie is 6.28in^2 The area of the second cookie is 18.84in^2

OpenStudy (gabylovesyou):

Since the area of a circle of radius r is A = π r², we see that the area is proportional to the square of the radius. So if the r increases from 2 to 6, i.e. increases by a factor of 6/2 = 3, then the area will increase by a factor of 3² = 9. So the price of the extra large cookie is 9 times that of the standard-sized cookie. so .20 x 9 = 1.80

OpenStudy (apoorvk):

yes, I misread it - your answer $1.80 is correct!! did you find out the area of the two cookies, and take ratio?

OpenStudy (gabylovesyou):

That is how i got it :D

OpenStudy (ash2326):

@Gabylovesyou great work:D Very nice:D

OpenStudy (apoorvk):

Your method is approximately 187.43% correct!! Brilliant!!

OpenStudy (anonymous):

nice work:)

OpenStudy (gabylovesyou):

:D lol thank you @ash2326 and @apoorvk and @jazy

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