The maximum volume of the box is 40√(10/27) cu in.
Here we see that volume is to be maximized
The surface area of the box is 40 sq in
Since the top lid is open, the surface area will be
lb + 2lh + 2bh = 40
Now, the length is equal to the breadth.
Let them be x in
Hence,
x² + 2xh + 2xh = 40
or, 4xh = 40 - x²
or, h = 10/x - x/4
Let f(x) = volume of the box
= lbh
Hence,
f(x) = x²(10/x - x/4)
= 10x - x³/4
differentiating with respect to x and equating it to 0 gives us
f'(x) = 10 - 3x²/4 = 0
or, 3x²/4 = 10
or, x² = 40/3
Hence x will be equal to 2√(10/3)
Now to check whether this value of x will give us the max volume, we will find
f"(2√(10/3))
f"(x) = -3x/2
hence,
f"(2√(10/3)) = -3√(10/3)
Since the above value is negative, volume is maximum for x = 2√(10/3)
Hence volume
= 10 X 2√(10/3) - [2√(10/3)]³/4
= 2√(10/3) [10 - 10/3]
= 2√(10/3) X 20/3
= 40√(10/27) cu in
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Complete Question
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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
2(x+2) > x+5
2x+4 > x+5
2x+4-x > x+5-x
x+4 > 5
x+4-4 > 5-4
x > 1
75%; so if we say there are 100 people in this city that means 60 of them live in houses and 40 of them live in apartments. Of the 60 living in houses 20% of them (12% of total population; 60 x .20) own their own business and those in apartments (40) 10% of them (4% of total pop.; 40 x .10) have their own business. In total 16 of the 100 people have businesses and 12 of those people live in houses so 12/16 equals .75. We multiply this by 100 to get the percent so there’s 75% that a person who owns a business in this city will also live in a house.
