A box with an open top is to be constructed by cutting equal-sized squares out of the corners of a 18 inch by 30 inch piece of cardboard and folding up the sides.
a) Let w be the length of the sides of the cut out squares. Determine a function V that describes the volume of the finished box in terms of w.
V(w) =
b) What width w would maximize the volume of the box?
w = inches
c) What is the maximum volume?
V = cubic inches
V(w) = (18-2w)(30-2w)
max V where dV/dw = 0
Oops. base * height is volume
V(w) = (18-2w)(30-2w)w
gfg
To determine the function V that describes the volume of the finished box in terms of w, we need to understand the construction of the box and how the measurements are related.
a) The construction involves cutting out equal-sized squares from each corner of the cardboard and folding up the sides to form the box. Let's assume that each side of the cutout square has a length of w.
The length of the cardboard piece is given as 30 inches, and the width is given as 18 inches. If we cut out squares of width w from each corner, the resulting length of the box would be 30 - 2w (as we remove w from each side) and the width would be 18 - 2w.
The height of the box is determined by the height of the cutout squares, which is w.
Therefore, the volume V of the box can be calculated as the product of the length, width, and height:
V(w) = (30 - 2w)(18 - 2w)(w)
b) To find the width w that maximizes the volume of the box, we need to determine the value of w that results in the maximum value of the function V(w). To find this, we can take the derivative of V(w) with respect to w and set it equal to zero.
Let's differentiate V(w):
V'(w) = [d(30 - 2w)/dw][(18 - 2w)w + (30 - 2w)dw/dw] + [d(18 - 2w)/dw][(30 - 2w)w + (18 - 2w)dw/dw] + [d(w)/dw][(30 - 2w)(18 - 2w)]
V'(w) = (-2)(18 - 2w)(w) + (30 - 2w)(18 - 2w) + (30 - 2w)(-2)(w) + (18 - 2w)(-2)(w) + (30 - 2w)(18 - 2w)(0)
V'(w) = -4w(18 - 2w) - 2(30 - 2w)(w) - 2(18 - 2w)(w) + 0
Simplifying this:
V'(w) = -8w^2 + 96w - 60w + 360 - 36w + 108
V'(w) = -8w^2 + 0w + 468
Setting V'(w) equal to zero:
-8w^2 + 468 = 0
Solving for w, we get:
w^2 = 468/8
w^2 = 58.5
w ≈ √58.5
w ≈ 7.65 inches
Therefore, the width w that maximizes the volume of the box is approximately 7.65 inches.
c) To find the maximum volume, we can substitute the value of w obtained in part b) back into the volume function V(w).
V ≈ V(7.65) ≈ (30 - 2(7.65))(18 - 2(7.65))(7.65)
Calculating this expression will give us the maximum volume in cubic inches.