The minimum surface area that such a box can have is 380 square
<h3>How to determine the minimum surface area such a box can have?</h3>
Represent the base length with x and the bwith h.
So, the volume is
V = x^2h
This gives
x^2h = 500
Make h the subject
h = 500/x^2
The surface area is
S = 2(x^2 + 2xh)
Expand
S = 2x^2 + 4xh
Substitute h = 500/x^2
S = 2x^2 + 4x * 500/x^2
Evaluate
S = 2x^2 + 2000/x
Differentiate
S' = 4x - 2000/x^2
Set the equation to 0
4x - 2000/x^2 = 0
Multiply through by x^2
4x^3 - 2000 = 0
This gives
4x^3= 2000
Divide by 4
x^3 = 500
Take the cube root
x = 7.94
Substitute x = 7.94 in S = 2x^2 + 2000/x
S = 2 * 7.94^2 + 2000/7.94
Evaluate
S = 380
Hence, the minimum surface area that such a box can have is 380 square
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It would help you to make a table because say your answer is 8 pieces is 1 minute and then the next question is how long wouls it take sam to cut 24 pieces you would use the table with you first answer and then multiple both numbers 3 times to give you your answer. Its really just a way to have more organized answers
Answer:
B : y=5/6cos(pi/30x)+9
Step-by-step explanation:
Edge 2020
Micah spent 28.10 dollars, Jeremy spent 34.82 dollars
41 times 0.15 equals 6.15. as you can't have 0.15 of a band, that means only 6 bands will compete.
