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: The answer is “10cm”
Step-by-step explanation: To get to this answer we first need to know the formula for find the hypotenuse which is a^2 + b^2 = c^2. A and B are the two sides in this case a and b are 8 cm and 6 cm. Then you plug in the values into the equation, it looks like this 8^2 + 6^ = c^2 once you solve you get 64 + 36 = c. When you add 64 and 36 you get 100 but their is one more step. You must find the square root of 100 which is 10. So your answer for the hypotenuse is “10cm”
Have a nice day!
Answer:
24
Step-by-step explanation:
Let x be the points he scored in his best game. x-8 would then represent each of his last 3 games.
(x-8)*3 = 2x
3x-24=2x
x=24
Answer:
z=t and y = 4y -- 6 for any t?
Well, seeing as how she mixed the clay, she has a total of 2 2/8 pounds clay (unsimplified). Although we don't know how many students she has in one class, with simple division of this number we can figure out how many student can finish the project. Removing the 2/8, we can already tell 2 students can do it, so that's a start. Now to divide the 2 pounds into 8ths. Each pound is enough for 8 students, so the 2 pounds totals to 16. Now add the extra 2 students, it's 18 total students. Hope I explained it well enough!
