The volume (in cubic inches) of a shipping box is modeled by V =2x³ -19x² +39x, where x is the length (in inches). Determine the values of x for which the model makes sense.

The values of x for which the model makes sense are 13/2 and 3

The equation that represents the volume of the box is given as:

Factor out x from the above equation

Expand the above equation, as follows

Factorize the above equation

Factor out x - 3

Equate to 0

Solve for x

or or

The value of x cannot be 0.

Hence, the values of x for which the model makes sense are 13/2 and 3

Read more about volumes at:

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The values of x for which the model is 0 ≤ x ≤ 3

Step-by-step explanation:

The given function for the volume of the shipping box is given as follows;

V = 2·x³ - 19·x² + 39·x

The function will make sense when V ≥ 0, which is given as follows

When V = 0, x = 0

Which gives;

0 = 2·x³ - 19·x² + 39·x

0 = 2·x² - 19·x + 39

0 = x² - 9.5·x + 19.5

From an hint obtained by plotting the function, we have;

0 = (x - 3)·(x - 6.5)

We check for the local maximum as follows;

dV/dx = d(2·x³ - 19·x² + 39·x)/dx = 0

6·x² - 38·x + 39 = 0

x² - 19/3·x + 6.5 = 0

x = (19/3 ±√((19/3)² - 4 × 1 × 6.5))/2

∴ x = 1.288, or 5.045

At x = 1.288, we have;

V = 2·1.288³ - 19·1.288² + 39·1.288 ≈ 22.99

V ≈ 22.99 in.³

When x = 5.045, we have;

V = 2·5.045³ - 19·5.045² + 39·5.045≈ -30.023

Therefore;

V > 0 for 0 < x < 3 and V < 0 for 3 < x < 6.5

The values of x for which the model makes sense and V ≥ 0 is 0 ≤ x ≤ 3.

step-by-step explanation:

through just a quick inspection, one can see that the graph of this cubing function starts in quadrant iii and ends in quadrant i.   when x = 0, y = 3, so the y-intercept is (0, 3).

if graphs are part of the problem statement, share them.   .

ummm seems that u didnt post the attachement !