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.
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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.
Answer:
It will take her 13 hours
7-2=5
Ans: 5 bags are small.
2w+2(10)=30
multioly the 2 and 10
2w+20=30
subtract 20
2w=10
divide by 2
w=5
plz rate as brainliest answer
