IQR = 6
First locate the median 
at the centre of the data arranged in ascending order. Then locate the lower and upper quartiles 
and 
located at the centre of the data to the left and right of the median.
Note that if any of the above are not whole values then they are the average of the values either side of the centre.
rearrange data in ascending order
15 16 ↓17 17 18 21 22 ↓23 25
↑
= 18
= 
= 16.5
= 
= 22.5
IQR = - = 22.5 - 16.5 = 6
The area enclosed is (2x)(600-2x).(2x)(600−2x)=1200x−4x2
In order to maximize we need the derivative to be equal to 0.y′=1200−8x=0
1200=8x
x=150
Therefore the sides for maximum area are 150*300.
<span>The area is: 45000</span>
