Answer:
All the sizes that satisfy ![kd^2 =144](https://tex.z-dn.net/?f=kd%5E2%20%3D144)
Step-by-step explanation:
To answer this question we first need to find the minimum wasted area of the dough.
Let us call the diameter of the cookie
, and
the length of the dough sheet, then the
number of cookies that fit into length
will be
![n = \dfrac{a}{d}](https://tex.z-dn.net/?f=n%20%3D%20%5Cdfrac%7Ba%7D%7Bd%7D)
and therefore, the number that will fit into the whole square sheet will be
![n^2 = \dfrac{a^2}{d^2}](https://tex.z-dn.net/?f=n%5E2%20%3D%20%5Cdfrac%7Ba%5E2%7D%7Bd%5E2%7D)
Since the area of each cookie is
![A = \pi \frac{d^2}{4}](https://tex.z-dn.net/?f=A%20%3D%20%5Cpi%20%5Cfrac%7Bd%5E2%7D%7B4%7D)
the area of n^2 cookies will be
,
which is the area of all the cookies cut out from the dough sheet; therefore, after the cutting, the area left will be
(1). ![\text{area left}= a^2-n^2\pi \frac{d^2}{4}](https://tex.z-dn.net/?f=%5Ctext%7Barea%20left%7D%3D%20a%5E2-n%5E2%5Cpi%20%5Cfrac%7Bd%5E2%7D%7B4%7D)
putting in the value of
we get
![a^2- \dfrac{a^2}{d^2}\pi \frac{d^2}{4}](https://tex.z-dn.net/?f=a%5E2-%20%5Cdfrac%7Ba%5E2%7D%7Bd%5E2%7D%5Cpi%20%5Cfrac%7Bd%5E2%7D%7B4%7D)
which simplifies to
area left = a^2( 1 - (π/4))
putting in a = 12 we get
area left = 30.902 in^2.
Going back to equation (1) we find that
a^2-n^2(πd^2/4) =30.902
12^2- n^2(πd^2/4) =30.902
and if we call k = n^2, we get
12^2- k(πd^2/4) =30.902
113.098 = k(πd^2/4)
simplifiying this gives
kd^2 = 144.
As a reminder, k here is the number of cookies cut from the dough sheet.
Hence, our cookie diameter must satisfy kd^2 = 144,<em> meaning larger the diameter of the cookies less of the should you cut out to satisfy the above equality. </em>