Let x be the side of the square and y the length of the box.
We must maximize the function f(x) = ???
What is the closed interval on that function?
What is the critical point?
What is f(x) of the critical point?
What is the maximum volume of a box satisfying the delivery's requirements?A parcel delivery............(Optimizat鈥?problem....Someone PLEASE help ASAP. Thank you)?
Corrected answer:
Dimensions are y * x * x, where x is the side of the square end.
y + 4x %26lt;= 112
If y + 4x %26lt; 112, then we could make the box bigger by expanding x or y,
so we'll use y + 4x = 112
y = 112 - 4x
volume = f(x) = (112 - 4x) x^2
= 112x^2 - 4x^3
Derivative = 224x - 12 x^2
Set that to 0
224 x = 12 x^2
224 = 12x
x = 18 2/3
Length = 112 - 72 = 40
Volume = 40 * 18 2/3 * 18 2/3 =~ 13009 cu in =~ 7.53 cu ft
Confirmation:
x...112-4x ... x^2 * (112 - 4x)
16.00 ... 48.00 ... 12,288.00
16.33 ... 46.67 ... 12,449.63
16.67 ... 45.33 ... 12,592.59
17.00 ... 44.00 ... 12,716.00
17.33 ... 42.67 ... 12,818.96
17.67 ... 41.33 ... 12,900.59
18.00 ... 40.00 ... 12,960.00
18.33 ... 38.67 ... 12,996.30
18.67 ... 37.33 ... 13,008.59 %26lt;%26lt; Max
19.00 ... 36.00 ... 12,996.00
19.33 ... 34.67 ... 12,957.63
19.67 ... 33.33 ... 12,892.59
20.00 ... 32.00 ... 12,800.00
No comments:
Post a Comment