If x represents the width of the poster (including borders), the area of the finished poster can be written as
.. a = x*(390/(x -10) +8)
.. = 8x +390 +3900/(x -10)
Then the derivative with respect to x is
.. da/dx = 8 -3900/(x -10)^2
This is zero at the minimum area, where
.. x = √(3900/8) +10 ≈ 32.08 . . . . cm
The height is then
.. 390/(x -10) +8 = 8 +2√78 ≈ 25.66 . . . . cm
The poster with the smallest area is 32.08 cm wide by 25.66 cm tall.
_____
In these "border" problems, the smallest area will have the same overall dimension ratio that the borders have. Here, the poster is 10/8 = 1.25 times as wide as it is high.
Per ticket cost $17.25
Per person conveyance fee $3.00
Percent increase in the cost of per ticket
17.39%
Given
81 red balls
Number of balls per row = Number of rows
Procedure
b = Number of balls
r = Number of rows
T = Total number of balls
b = r
![\begin{gathered} T=b\cdot r \\ Given\text{ b=r} \\ T=b^2 \\ b=\sqrt[]{T} \\ b=\sqrt[]{81} \\ b=9 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20T%3Db%5Ccdot%20r%20%5C%5C%20Given%5Ctext%7B%20b%3Dr%7D%20%5C%5C%20T%3Db%5E2%20%5C%5C%20b%3D%5Csqrt%5B%5D%7BT%7D%20%5C%5C%20b%3D%5Csqrt%5B%5D%7B81%7D%20%5C%5C%20b%3D9%20%5Cend%7Bgathered%7D)
The answer is:
Number of rows = 9
Number of balls = 9
5/8=x/100
x=(5/8)(100)=500/8
x=62.5%
