SOLUTION: Bruslan is the architect in charge of designing a new elementary school. He has been told that each rectangular classroom should include 25m^2 of common space, plus 1.50m^2 per stu

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Question 1150998: Bruslan is the architect in charge of designing a new elementary school. He has been told that each rectangular classroom should include 25m^2 of common space, plus 1.50m^2 per students for desks and chairs. If the maximum number of students in the class is 22, what is the least possible perimeter that the classroom design could have?
Found 2 solutions by ankor@dixie-net.com, ikleyn:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Bruslan is the architect in charge of designing a new elementary school.
He has been told that each rectangular classroom should include 25m^2 of common space, plus 1.50m^2 per students for desks and chairs.
If the maximum number of students in the class is 22, what is the least possible perimeter that the classroom design could have?
:
Find the total area of the class room.
25 + 22(1.5) = 58 sq/m
:
the area:
L * w = 58
replace w with (58/L)
w = 58%2FL:
:
the perimeter
P = 2L + 2w
P = 2L + 2(58/L}
Graph this equation
+graph%28+300%2C+200%2C+-10%2C+20%2C+-10%2C+50%2C+2x%2B2%2858%2Fx%29%2C+30.5%29+
Graph shows Length for minimum perimeter occurs at 7.5 meters
therefore
58%2F7.5 = 7.73 meters is the width
;
Min perimeter
P = 2(7.5) + 2(7.73)
P = 30.5 meters (green line)

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.

            The correct solution is below. 


You determine first the area of the class room for 22 students

    Area = 25 + 22*1.5 = 58 square meters.


A rectangle which provides the minimum perimeter at given area, is THE SQUARE with the side length equal to square root of the area.


In this case, the classroom with minimal perimeter is the square with the side length of

    sqrt%2858%29 = 7.616 meters.    


The least possible perimeter is then  4*7.616 = 30.464 meters (approximately).       ANSWER

Solved.

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If you want to see many other similar optimization problems,  look into my lesson
    - Calculus optimization problems
in this site.

Specifically,  for the given problem,  its precise analogue is the  Problem 3  of the referred lesson.