Let us say room has length x ,width w and area A.
Perimeter `= 2(x+w)`
`400 = 2(x+w)`
`w = 200-x`
It is given that a student occupies 2.5m^2 of space. When the area of the room maximizes then the student will also maximize.
If the amount of students are P then;
`P = A/2.5`
`P = (x*w)/2.5`
`P = (x*(200-x))/2.5`
`P = (200x-x^2)/2.5`
For maximum or minimum number of students` (dP)/dx = 0`
`(dP)/dx = 1/2.5(200-2x)`
When `dP/dx = 0`
`1/2.5(200-2x) = 0`
`x = 100`
If P has a maximum at x = 100 then `[(d^2P)/dx^2]_(x = 100)` is a negative value.
`(d^2P)/dx^2 = -2<0`
So we have a maximum for P.
Maximum area of the room `= 100x100m^2`
Maximum students `= (100xx100)/2.5 = 4000`
So we can have maximum 4000 students in the room.
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