The formula for the volume of the box and the maximum volume are presented as follows;
(a) V(x) = 4·x³ - 44·x² + 112·x
(b) 0 < x < 4
(c) Please find attached the graph of the volume function, created with MS Excel
(d) 82.98 cubic inches
The steps used to arrive at the above values are as follows;
(a) The dimensions of the box formed by cutting out squares from the corners are;
Width, W = 8 - 2·x
Length, L = 14 - 2·x
Height, H = x
Volume, V(x) = W × L × H
W × L × H = (8 - 2·x) × (14 - 2·x) × x
(8 - 2·x) × (14 - 2·x) × x = 4·x³ - 44·x² + 112·x
V(x) = 4·x³ - 44·x² + 112·x
(b) The values of x for which the formula makes sense can be obtained from the values of the expressions for the length, width and height that have positive values. therefore;
Width, W ; 8 - 2·x > 0
Length, L; 14 - 2·x > 0
Height, H; x > 0
8 - 2·x > 0, indicates; 8 > 2·x
2·x < 8
x < 8/2
x < 4
Similarly from the inequality, 14 - 2·x > 0, x < 7
4 < 7, therefore, 0 < x < 4 satisfies the inequalities, x < 7, and x > 0. The values of x for which the formula makes sense is; 0 < x < 4
(c) The graph of the volume function can be constructed by finding the values of V(x), for values of x from 0 to 4, as follows;
V(0) = 4 × 0³ - 44 × 0² + 112 × 0 = 0
V(1) = 4 × 1³ - 44 × 1² + 112 × 1 = 72
V(2) = 4 × 2³ - 44 × 2² + 112 × 2 = 80
V(3) = 4 × 3³ - 44 × 3² + 112 × 3 = 48
V(4) = 4 × 4³ - 44 × 4² + 112 × 4 = 0
Please find attached the graph of the volume function, V(x), created using MS Excel using the ordered pair of points generated by the function
(d) The volume function, V(x) indicates;
V(x) = 4·x³ - 44·x² + 112·x
Therefore;
V'(x) = d(4·x³ - 44·x² + 112·x)/dx
V'(x) = 12·x² - 88·x + 112
The maximum volume can be found using the equation, V'(x) = 0, which corresponds to the point where the slope of the graph is zero, which are at the lowest or highest point on the graph
Therefore, at the maximum point;
V'(x) = 0
12·x² - 88·x + 112 = 0
4·(3·x² - 22·x + 28) = 0
3·x² - 22·x + 28 = 0
x = (22 ± √(22² - 4×3×28))/(2×3)
x = (22 ± 2×√(37))/6
x = (11 ± √(37))/3
x ≈ 5.69 and x ≈ 1.639
The maximum volume of the box is; V(x) = 4·x³ - 44·x² + 112·x
V(1.639) = 4 × 1.639³ - 44 × 1.639² + 112 × 1.639
4 × 1.639³ - 44 × 1.639² + 112 × 1.639 = 82.98
The maximum volume is about 82.98 cubic inches
The complete question found through search can be presented as follows;
An open top box is to be constructed from 8 in. by 14 in. rectangular sheet of tin by cutting out squares of equal size at each corner, then folding up the resulting flaps. Let x denote the length of the side of each cut-out square. Assume negligible thickness.
(a) Find the formula for the volume, V, of the box as a function of x. V(x) =
(b) For what values of x does the formula from part (a) make sense in the context of the problem?
(c) On a separate piece of paper, sketch a graph of the volume function.
(d) What, approximately, is the maximum volume of the box?