A parcel delivery............(Optimizat鈥?problem....Someone PLEASE help ASAP. Thank you)?

A parcel delivery service will deliver a package only if the length plus the girth (distance around) does not exceed 112 inches. Find the maximum volume of a rectangular box with square ends that satisfies the delivery company's requirements.



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

things to do in miami

virtual villagers 5