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About Alon Mandes
Expertise
Kind of questions I can answer : Limits, Derivatives, Integration, Implicit functions, continuousity, differentiation ,Extremum problems, Lagrange multipliers, Gradients, Surface integrals, Multi variables functions ,Multi variables Integrals,Complex variables ,Complex functions, Curves, Trajectory integrals & Vector analyse,Divergence,Rotor & word problems. Kind of question I can't answer : Economics,Combinatorics,infinite series & convergence ,Statistics & Probabilities .

Experience
1. I'm a team member of mathnerds (math site for answering questions) 2. I'm a team member in the Student's Union of the Technion, helping students who have problems in mathematics. 3. 2 years of experience as a math teacher in college. 4. I give free homework help for high school students in Mathematics & Physics. 5. I teach part time in collage the subjects : "Digital Signal Processing" , "Random Signals & Noise" , "Complex Functions".

Organizations
Hi-Tech company : GSM4VOIP ; job possition : Algorythm developer.

Education/Credentials
M.A in Mathematics & Bs.c in Electronics.

You are here: Experts > Teens > Homework/Study Tips > Calculus > Optimization Problem

Calculus - Optimization Problem

Expert: Alon Mandes - 11/7/2009

Question
Can you show me how to work through this problem? THANK YOU!
5. (1 pt) A parcel delivery service will deliver a package only
if the length plus the girth (distance around) does not exceed 100
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) = over the interval
[a,b] with a = and b = .
Note that the volume of the box is maximum when the distance
around plus the girth is maximum.
We find that f (x) has a unique critical point in the interval (a,b)
at Xo = . The values of the function at the end-points and
critical point are f (a) = , f (x0) = and f (b) = .
Therefore the maximum volume of a box satisfying the delivery’s
requirements is .

Answer
If x is the side of the square and y the length of the box, then :
Volume=yx²2 & girth=y+4x≤100 .
f(x)=yx²=(100-4x)x² . The interval will be [a,b] where : a=x=1 then y=96 & y=1 then b=x=99/4 . Therefore:
f'(x)=-4x²+2x(100-4x) . f'(x)=0 --> -4x²+2x(100-4x)=0 --> 2x=100-4x --> xo=100/6=16.66
So yo=100-4xo=100-4*16.66=33.33
f''(x)=-8x+2(100-4x)-8x=100-4x-16x=100-20x . f''(16.66)=-233.33 So it's maximum.
f(a)=f(1)=1²*(99/4)=24.75
f(xo)=f(16.66)=33.33*16.66²=9159
f(b)=f(99/4)=(99/4)²*1=612.56
Therefore the maximum volume of a box satisfying the delivery’s
requirements is 9159 .

Alon.

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