An open box with a square base is to bemade to organize items in the trunk of a car. It is planned to use an area of 16m square of flexible plastic forthis box. What are the dimensions of the box for max volume?
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To find the dimensions of the box that maximize the volume, we need to understand the relationship between the variables involved. Let's assume the length of one side of the square base is "x", and the height of the box is "h".
We know that the area of the square base is given by the formula: Area = length x width.
Since it's a square base, the length and width are both equal to "x". Therefore, the area of the square base is x^2.
Now, let's consider the surface area of the box. The box consists of five sides: the square base and four equal rectangular sides.
The area of each rectangular side is given by the formula: Area = length x width. Since the length is "h" and the width is "x", the area of each rectangular side is hx.
Therefore, the surface area of the box is: 2(x^2) + 4(hx).
According to the problem, the total area of flexible plastic available is 16m^2.
So, we can write the equation: 2(x^2) + 4(hx) = 16.
Now, we need to express the volume of the box in terms of one variable so that we can find its maximum value. The volume of a box is given by the formula: Volume = length x width x height.
In this case, the volume is V = x^2h.
To eliminate "h" from the equation, we can isolate "h" from the surface area equation mentioned earlier: h = (16 - 2x^2) / (4x).
Now, substitute this value of "h" into the volume equation to get: V = x^2((16 - 2x^2) / (4x)).
Simplifying further, V = (4x^2 - 0.5x^4) / 2.
Now, to find the maximum volume, we need to find the critical points of this function. To do that, we differentiate the volume function with respect to "x" and set it equal to zero.
dV/dx = 8x - 2x^3 = 0.
Factor out x to get: x(8 - 2x^2) = 0.
Setting each factor equal to zero, we get two critical points: x = 0 and 2 = x^2.
Since the length of the box cannot be zero, we consider x^2 = 2.
So, x = ±√2.
Since x represents the length of a side, it cannot be negative. Therefore, x = √2.
Now, substitute this value of "x" back into the equation h = (16 - 2x^2) / (4x) to find the corresponding height.
h = (16 - 2(√2)^2) / (4√2) = (16 - 4) / (4√2) = 12 / (4√2) = 3 / √2 = (3√2) / 2.
Therefore, the dimensions of the box that maximize the volume are: length = width = √2 meters and height = (3√2) / 2 meters.