(a) The function that models the volume V of the box is: V(x) = (30 - 2x)(12 - 2x)(x)
(b) The values of x for which the volume is greater than 250 in³ are in the interval (0, 2.5).
(c) The largest volume that such a box can have is approximately 649.5 in³.
(a) Find a function that models the volume V of the box.
The volume of a box is calculated by multiplying the length, width, and height of the box. In this case, the length and width of the box will be reduced by 2x, since we are cutting out squares of side x at each corner. The height of the box will be equal to x.
Therefore, the function that models the volume V of the box is:
V(x) = (30 - 2x)(12 - 2x)(x)
(b) Find the values of x for which the volume is greater than 250 in³.
To find the values of x for which the volume is greater than 250 in³, we can set up the following inequality:
V(x) > 250
Substituting the function for V(x), we get:
(30 - 2x)(12 - 2x)(x) > 250
Expanding the left-hand side of the inequality, we get:
360x - 192x² + 48x³ > 250
Subtracting 250 from both sides of the inequality, we get:
48x³ - 192x² + 360x - 250 > 0
We can use a graphing calculator or software to solve this inequality. The solution is:
0 < x < 2.5
Therefore, the values of x for which the volume is greater than 250 in³ are in the interval (0, 2.5).
(c) Find the largest volume (in in³) that such a box can have.
To find the largest volume that such a box can have, we can set the derivative of the function V(x) equal to zero and solve for x. The derivative of V(x) is:
V'(x) = 48x² - 72x + 360
Setting V'(x) equal to zero, we get:
48x² - 72x + 360 = 0
Dividing both sides of the equation by 24, we get:
2x² - 3x + 15 = 0
Using the quadratic formula, we get the following solutions:
x = 0.625 or x = 2.375
Since x cannot be equal to zero, the largest volume that such a box can have is achieved when x is equal to 2.375.
Substituting this value into the function V(x), we get the following volume:
V(2.375) ≈ 649.5 in³
Therefore, the largest volume that such a box can have is approximately 649.5 in³.