Postal regulations specify that a parcel sent by priority mail may have a combined length and girth of no more than 132 in. find the dimensions of a rectangular package that has a square cross section and largest volume that may be sent by priority mail. hint: the length plus the girth is 4x + l.

v = 11264 in^3

Step-by-step explanation:

4 x + l =132

l=132- 4 x

v= h x^ 2

 = x ^ 2 (132 - 4 x )

= 132 x ^ 2 - 4 x^3

 = 4 ( 33 x^2 - x ^3)

d v/d x = 4 (66 x - 3 x^2)

         = 12 x ( 22 - x)

d v/d x = 0

when x = 0  or 22

So

h = 132 - 4 *22 = 44

So if the package is  44*16*16

v = 11264 in^3

So  4 x+l >= 132 But it should be minimum if 4 x+l contain smaller value

10a^3+277a^2-132a-792

step-by-step explanation:

(2a+11)(5a-12)(foil)

=10a^2-24a+55a-132 =10a^2+31a-132

(10a^2+31a-132)(a+6)

=10a^3+60a^2+31a^2+186a^2-132a-792 =10a^3+277a^2-132a-792

i hope this is what you are looking for : )