Expert: Alon Mandes - 11/9/2009

Question
QUESTION: Any chance you could look at another optimization problem?  I tried to attach the PDF document but this site doesn't allow me to.  I copied it below: I am totally stuck on this one too.  Any support that you can provide is most appreciated.  (if there are any decimals......I have to round to 5 places).
I just don't know where to begin with these!!!  
Sincerely,
Mikel  

4. (1 pt) A box is to be made out of a 8 by 18 piece of cardboard.
Squares of equal size will be cut out of each corner, and
then the ends and sides will be folded up to form a box with
an open top. Find the length L, width W, and height H of the
resulting box that maximizes the volume. (Assume that W less that or equal to L).
Let x be the side of the squares removed. To solve this problem
we should maximize the function V(x)= ____?___over the
interval [a,b] with a =____?___ and b =____?____ .
The function V(x) has a unique critical point in (a,b) at x0 = ____?____. The values of the function at the end points and critical point are V(a) = ____?____, V(x0) = ____?____and V(b) =___?____ .
Thus, the box with maximum volume has dimensions
L =___?___ ’W =___?___ and H =___?___




Let x be the side of the squares removed. To solve this problem
we should maximize the function V(x)= over the
interval [a,b] with a = and b = .
The function V(x) has a unique critical point in (a,b) at x0 =
. The values of the function at the end points and critical
point are
V(a) = , V(x0) = and V(b) = .
Thus, the box with maximum volume has dimensions
L = ’W = and H =
5. (1 pt) A parcel delivery service will deliver a package


ANSWER: Our cardboard has dimensions of : 8X18 .
Our function is V(x)=(18-2x)(8-2x)x=(144-36x-16x+4x²)x=144x-48x²+4x³ .
The interval will be : x=a=0 & x=b=4
V'(x)=144-96x+12x² . V'(x)=0 --> x1=6   & x2=2 . To know which max & which min , we find :
v''(x)=-96+24x . V''(2)=-96+48=-48 therefore max . V''(6)=-96+24*6= 48 therefore min .
So xo=2.

V(a)=v(0)=(18-2*0)(8-2*0)*0=0
V(xo)=V(2)=(18-2*2)(8-2*2)*2=112
V(b)=V(4)=(18-2*4)(8-2*4)*4=0

L=18-2*4=10   &  W=8-2*2=4   & H=2 .

Alon.

---------- FOLLOW-UP ----------

QUESTION: here are the results after I plug it in to webwork: (this is tough)
Entered Answer Preview Result
v(x) = 144*x-48*(x^2)+4*(x^3)  incorrect
interval (a) = 0  correct
interval (x) = 4  correct
Xo = 2  incorrect
V(a) = 0  correct
V(Xo)= 2  incorrect
V(b) = 0  correct
L = 10  incorrect
W = 4  incorrect
H = 2  incorrect  

Answer
sorry, The v(x) should be : V(x)=144x-54x²+4x³ . So, v'(x)=144-108x+12x² .
v'(x)=0 : 144-108x+12x²=0 : x1=7.3723 & x2=1.6277 .
v''(x)=-108+24x --> v''(1.627)=-69 This is max point xo=1.627.
v(xo)=144xo-54xo²+4xo³=108.5706

L = 18-2*1.627= 14.7460  
W = 8-2*1.627= 4.7460  
H = 1.6277

Alon.