Answer:
4 in × 4 in × 8 in or
6.47 in × 6.47 in × 3.06 in
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