A company is manufacturing an open-top rectangular box. They have 30 cm by 16 cm sheets of material. The bins are made by cuttin
1 answer:
Answer:
a. (30 - 2x)(16 - 2x)x. b. length 21.2 cm , width = 7.2 cm and height x = 4.4 cm
Step-by-step explanation:
a. Let x be the side of the squares to be cut from each corner. Since we have two corners on each side, the length of the resulting box from the 30 cm by 16 cm sheet is L = 30 - 2x. Its breadth is B = 16 - 2x. The height of the resulting box is x. So its volume V = LBx = (30 - 2x)(16 - 2x)x.
b. To find the maximum value of V, we differentiate V with respect to x and equate it to zero.
So, dV/dx = 0
d(30 - 2x)(16 - 2x)x/dx = 0
-2(16 - 2x)x + (-2)(30 - 2x)x + (30 - 2x)(16 - 2x) = 0
32x + 4x² - 60x - 4x² + 480 - 60x - 32x + 4x² = 0
4x²- 120x + 450 = 0
x²- 30x + 112.5 = 0
Using the quadratic formula, we find x. So, with a = 1, b = - 15 and c = 112.5,
x ≅ 25.61 or 4.4
V = (30 - 2x)(16 - 2x)x
Substituting the values of x into L and B, we have )x
L = (30 - 2x) = (30 - 2(25.61)) = 30 - 51.22 = -21.22
B = (16 - 2x) = (16 - 2(25.61)) = 16 - 51.22 = -35.22
Since L and B cannot be negative, we use the other value for x = 4.4, So
L = (30 - 2x) = (30 - 2(4.4)) = 30 - 8.8 = 21.2
B = (16 - 2x) = (16 - 2(4.4)) = 16 - 8.8 = 7.2
So V = LBx = 21.2 × 7.2 × 4.4 = 671.62 cm³
You might be interested in
x = -4 OR x = 6
The correct options are B and G
Answer:
B. (6, 1)
Step-by-step explanation:
A graph of the solution space shows that the point (6, 1) is the only one that satisfies both inequalities.
___
<em>Verification</em>
3(6) +4(1) = 18 +4 = 22 < 24
3(6) -4(1) = 18 -4 = 14 > 12
_____
<em>Notes on the Solution</em>
The graphing calculator used here only recognizes variables x and y. Since g is the first variable in the ordered pair (g, z) we use x to represent g in the equations that are graphed. Similarly, y is used to represent z in the graphed equations.
You graph an inequality by graphing it as though it were an equation, then choosing the appropriate half-plane (one side of the line or the other) to shade as the solution space. Where the shadings from the two inequalities overlap, both are satisfied by points in that area.
In the first inequality, x and y values that are less than those on the line will satisfy the inequality, so the shading is below the line.
In the second inequality, x values greater than those on the line will satisfy the inequality, so the shading is to the right of the line.
In both cases, the line is graphed as a dashed line. This is because the points on the line do not satisfy the < or > conditions, so are not part of the solution set. (If one or both conditions were ≤ or ≥, then the corresponding line would be solid, and the points on the line would be part of the solution.)
