According to US Postal regulations, a rectangular shipping box must satisfy the

Question

According to US Postal regulations, a rectangular shipping box must satisfy the inequity "Length + Girth lessthanorequalto 130 inches" for parcel post and "Length + Girth lessthanorequalto 108 inches" for other services. Let's assume we have a closed rectangular box with a square face of side length x as drawn below. The length is the longest side and is clearly labeled. The girth is the distance around the box in the other two dimensions so in our case it is the sum of the four sides of the square, 4x. (a) Assuming that we'll be mailing a box via Parcel Post where Length + Girth = 130 inches, express the length, l, of the box in terms of x and then express the volume, V, of the box in terms of x. l(x) = V(x) = (b) Find the dimensions of the box of maximum volume that can be shipped via Pared Post. width in height in length in (c) Repeat parts (a) and (b) if the box is shipped using "other services". length l(x) = volume V(x) = width in height in length in

Explanation / Answer

Let the dimensions of the box be x by y by z
It would be so very nice if we could eliminate all but one unknown... but we cannot. I shall assume that you have had 2 and 3 variable calculus
and understand A /y and A /x...
Here we go :
the total girth + length = 2x +2y + z = 108 ( or less )
so x = 108 - 2x -2y
The volume is :
V = lwh = x y z = x y (108 - 2x -2y ) still two variables ...

V /x = Vx = 108y -4xy - 2y^2 and V /y = Vy = 108x - 2x^2 -4xy
and differentiating again , to find the second derivatives :
Vxx = -4y and Vyy = -4x
and any maximum should have these two EQUAL
so where is -4x = -4y ? when x = y
Aha ! a square cross section.
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PS You could have also arrived at this by " Knowing " that the maximum area of a rectangle is when its a square : x = y
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So now , if x = y , then the volume becomes :
V = x^2 ( 108 - 2(2x) = 108x^2 -4x^3
now dV/dx = V' = 216x -12x^2 =x(216-12x)
so by the Zero Product Property
Either x = 0 ( Minimum ) or 216 -12x = 0( Max )
so x = 216 /12 = 18
and y also = 18 , and z = 108 - 2(2(x)) = 108 - 72 = 36

so the dimensions of the largest box are :
18 by 18 by 36 <----- and this gives a volume of 11644 cubic units

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