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ale4655 [162]
3 years ago
14

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 t

hat the sum of the length of the length of its edges is equal to 64 inches, what should be the box dimensions?
Mathematics
1 answer:
sergij07 [2.7K]3 years ago
7 0

Answer:

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

 <u>|      4  -16  -64 </u>

   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    

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