The largest possible volume of the given box is; 96.28 ft³
<h3>How to maximize volume of a box?</h3>
Let b be the length and the width of the base (length and width are the same since the base is square).
Let h be the height of the box.
The surface area of the box is;
S = b² + 4bh
We are given S = 100 ft². Thus;
b² + 4bh = 100
h = (100 - b²)/4b
Volume of the box in terms of b will be;
V(b) = b²h = b² * (100 - b²)/4b
V(b) = 25b - b³/4
The volume is maximum when dV/db = 0. Thus;
dV/db = 25 - 3b²/4
25 - 3b²/4 = 0
√(100/3) = b
b = 5.77 ft
Thus;
h = (100 - (√(100/3)²)/4(5.77)
h = 2.8885 ft
Thus;
Largest volume = [√(100/3)]² * 2.8885
Largest Volume = 96.28 ft³
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Answer:
x=4
Step-by-step explanation:
Remove the parentheses and turn the 4-2x and make it into 4+2x candle the equal term -4 and should be left with x+6=2+2x then move the variable to the left and change the sign so it would be x-2x=2-6 and collect like terms so it would be -x=2-6 and then calculate that and get -x= -4
(sorry its alot)
The answer is -18. Hope this helps.
To find the interquartile range, you will list the data that is presented in the stem and leaf plot.
Find the median of the data (30.5)
Find the median of the lower half and the median of the upper half.
Subtract these two values.
The data are <u>20</u>, 25, 30, 30, 31, 40, 41, <u>49</u>.
27.5 40.5
40.5-27.5 = 13
The interquartile range is 13.
Answer:
the median would be 3
Step-by-step explanation:
