Mathamatics

(Full name (Professor (Course (18 December 2006 (Title

a ) The U .S . Postal Service will accept a box for domestic shipment only if the sum of its length and girth (distance around ) does not exceed 108 in (The last time I called the post office this was still true Suppose that the box has a square end ( See in the Picture . What dimensions will give a box of maximum volume

Let L length of box

x side of square end

108 length girth

108 L 4x

L 108 – 4x p [banner_entry_middle]

Volume f (x (length (width (height (L (x (x (108 – 4x (x2 108×2 – 4×3

The function is therefore : f (x 108×2 – 4×3

Domain of x : [0 , 108]

We find the critical numbers (first derivative ) to know the relative extrema

f (x 108×2 – 4×3

f (x (2 (108 )x – (4 (3 )x2

We set the value to zero to find the critical numbers

0 (2 (108 )x – (4 (3 )x2

0 216x – 12×2

0 x (216 – 12x

Critical numbers : x 0 or 18

Although x 0 is in the domain , the side of a box can ‘t be zero so x could only be 18

x 18

To check if at x 18 , the y value is a relative maximum , we find the second derivative

f (x 108×2 – 4×3

f (x 216x – 12×2

f (x 216 – (2 (12 )x 216 – 24x

We substitute 18 to the equation 216 – (24 (18 -216 ( The value is negative , which means that at the point where x 18 , the function value is at a relative maximum . It is also the absolute maximum value because the values for the side of the box x 0 and x 108 can ‘t be true

Now that we have x , we now find the length of the box

L 108 – 4x 108 – (4 (18 36

Thus , the dimensions of a box of maximum value are : length x width x height 36 x 18 x 18

b ) Suppose that instead of having a box with square ends you have a box with square sides (See the picture ) What dimensions will give the box of largest volume

Let L length of box

X other side

108 length girth

108 L (2L 2x

108 3L 2x

L (108 – 2x / 3

Volume f (x (length (width (height (L (x (L L2x

Domain of x : [0 , 108]

To find the critical numbers , we find the first derivative of the function and equate it with 0

f ` (x (x (2

0 (27 – 2 /3x )2 – 4 /3x (27 – 2 /3x

0 272 – 36x 4 /9×2 – 36x – 8 /9×2

0 272 – 72x 4 /3×2

x 40 .5 , 13 .5

We find the second derivative of the function to find out which of the two is at a relative maximum

f ` (x 362 – 96x 4 /3×2

f (x (2 (4 /3 )x – 96 8 /3x – 96

Case 1 : x 40 .5

f (x (8 /3 (40 .5 ) -… [banner_entry_footer]