We want to build a box with square base of side x and height and in such a way that the volume is 128 cubic inches. if we know that the sum of the length of the length of its edges is equal to 64 inches, what should be the box dimensions?

4 in × 4 in × 8 in or  

6.47 in × 6.47 in × 3.06 in

Step-by-step explanation:

Data:

(1)                  V = 128 in³

(2)                  l = w = x

(3) 4(l + w + h) = 64 in  (There are 12 edges)

Calculation:

The formula for the volume of the box is

(4)                 V = lwh

(5)              128 = x²h          Substituted (1) and (2) into (4)

(6)                  h = 128/x²     Divided each side by x²

          l + w +h = 16            Divided (1) by 4

          x + x + h = 16           Substituted (2) into 6

(7)          2x + h = 16           Combined like terms

     2x + 128/x² = 16           Substituted (6) into (7)

        2x³ + 128 = 16x²        Multiplied each side by x²

2x³ - 16x²+ 128 = 0            Subtracted 16x² from each side

    x³ - 8x² + 64 = 0           Divided each side by 2

According to the Rational Zeros theorem, a rational root must be a positive or negative factor of 64.

The possible factors are ±1, ±2, ±4, ±8, ±16, ±32, ± 64.

After a little trial-and-error with synthetic division (start in the middle and work down) we find that x = 4 is a zero.              

4|1   -8     0   64

 |      4  -16  -64

   1  -4  -16      0

So, the cubic equation factors into (x - 4)(x² - 4x + 16) = 0

We can use the quadratic formula to find that the roots of the quadratic are

x = 2 - 2√5 and x = 2+ 2√5

We reject the negative value and find that there are two solutions to the problem.

x = 4 in and x = 2 + 2√5 ≈ 6.472 in

Case 1. x = 4 in

h = 128/x² = 128/4² = 128/16 = 8 in

The dimensions of the box are 4 in × 4 in × 8 in

Also, 4(l + w + h) = 4( 4 + 4 + 8) = 4 × 16 =  64 in

Case 2. x = 6.472 in

h = 128/x² = 128/6.472² = 128/41.89 = 3.056 in

The dimensions of the box are 6.47 in × 6.47 in × 3.06 in

Also, 4(l + w + h) = 4( 6.47 + 6.47 + 3.06) = 4 × 16.00 =  64 in

The two solutions are

(a) 4 in       × 4 in      × 8 in

(b) 6.47 in × 6.47 in × 3.06 in    

its 50 miles away from the starting point