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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Given:
The function is
To find:
The asymptotes and zero of the function.
Solution:
We have,
For zeroes, f(x)=0.
Therefore, zero of the function is 0.
For vertical asymptote equate the denominator of the function equal to 0.
Taking square root on both sides, we get
So, vertical asymptotes are x=-4 and x=4.
Since degree of denominator is greater than degree of numerator, therefore, the horizontal asymptote is y=0.
Answer:
Step-by-step explanation:
400×10=4000
400×2=800
50×10=500
50×2=100
2×10=20
2×2=4
4000+800+500+100+20+4=5,424
Check answer by multiplying 452 and 12
452×12=5,424
Answer:
the answer would be 13
Step-by-step explanation:
26/2=13
