Suppose that postal requirements specify that parcels must have length plus girth at most 78 inches. Consider the problem of finding the dimensions of the square ended rectangular package of greatest volume that is mailable. A rectangular prism has length labeled X, width labeled h, and height labeled x. Required:
a. Express the length plus the girth in terms of x and h.
b. Determine the objective and constraint equations.
c. What is the constraint equation?

A. 4x + h

B. 4x + h = 78

V = x²h

Step-by-step explanation:

1. The length plus girth

X +x+x+x +h

= 4x +h

2. Constraint equation

4x + h = 78 ----(1)

Objective function

Volume = v = x*x*h

V = x²h ----(2)

3. Quantity as function of x

From equation 1 in answer part b

4x+h = 78

We make h subject

h = 78-4x

We put values of h in equation 2 in part b

V = x²h

V = x²(78-4x)

V = 78x²-4x³

To get values of x and h

V = 78x²-4x³

Dv/dx = 0

= 156x - 12x²

156x = 12x²

Divide through by 12x

156x/12x = 12x²/12x

x = 13

h = 78-4x

h = 78 - 4(13)

h = 78 - 52

h = 26 inches

6mm^3 or .6cm^3

      216

      /   \

  36     6

  /   \     /

6     6   6

factor trees are nice.

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752