A box must have base x cm by 2x cm. what are the dimensions of the box of maximum volume if its length+width+depth<140cm
2 x^2 h = v
h = 140-3x
v =2 x^2 (140-3x)
v = 280 x^2 - 6 x^3
dv/dx = 0 = 560 x - 18 x^2
x(560 - 18x) = 0
18 x = 560
x = 31.1
2x = 62.2
h = 140 - 3(31.1) = 46.7
To find the dimensions of the box with the maximum volume, we need to formulate an equation for the volume of the box and then optimize it.
Let's assume the dimensions of the box are x cm by 2x cm by y cm (length, width, and depth, respectively).
The volume of a rectangular box is given by the formula: V = l * w * h, where V is the volume, l is the length, w is the width, and h is the height.
In this case, the length is x cm, the width is 2x cm, and the height is y cm, so the volume of the box is given by:
V = x * 2x * y = 2x^2y
We also have the constraint that the sum of the length, width, and depth should be less than 140 cm:
x + 2x + y < 140
Now, let's simplify the constraint equation:
3x + y < 140
To find the dimensions of the box with the maximum volume, we need to optimize the volume function (2x^2y) subject to the constraint equation (3x + y < 140).
To do this, we can use calculus. We need to determine the critical points of the volume function and check which ones satisfy the constraint equation.
1. First, let's find the partial derivatives of the volume function with respect to x and y:
dV/dx = 4xy
dV/dy = 2x^2
2. Next, set the partial derivatives to zero to find the critical points:
4xy = 0 => xy = 0
2x^2 = 0 => x^2 = 0
From xy = 0, we can deduce that either x or y must be equal to zero. However, if any dimension is zero, then the volume would also be zero, which is not the maximum volume we are seeking.
Therefore, let's focus on the equation x^2 = 0, which implies x = 0.
3. Substitute the critical points into the constraint equation:
3x + y < 140
3(0) + y < 140
y < 140
Since y can be any positive value less than 140, we can conclude that there is no upper bound on the value of y.
4. Finally, consider the lower bound of y. We know that y must be greater than zero because we cannot have any dimension as zero.
Therefore, the dimensions of the box of maximum volume are as follows:
x = 0 (minimum possible value)
y > 0 (minimum value greater than zero)
width: 2x = 2 * 0 = 0 cm
length: x = 0 cm
depth: y (positive value greater than zero)
This means that the box would have zero width and length, with the only constraint being that the depth (y) must be a positive value greater than zero.
Please note that this result may seem counterintuitive or impractical in real-world scenarios as the box would essentially be a flat surface. However, based on the given conditions and calculations, this is the answer to the problem.