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
To find the dimensions of the box with maximum volume, we need to apply optimization principles. Let's break down the problem step by step:
1. Begin by defining the variables: Let's consider x as the length of the base of the box in centimeters. Therefore, the width will be 2x centimeters. The height of the box will be h centimeters.
2. Determine the volume of the box: The volume of a box is given by V = lwh, where V represents volume, l represents length, w represents width, and h represents height. In this case, the formula becomes V = x * 2x * h = 2x^2 * h.
3. Identify the constraint: The constraint provided states that the sum of the length, width, and depth (height) of the box must be less than 140 cm. Mathematically, this can be represented as x + 2x + h < 140.
4. Rewrite the constraint equation in terms of a single variable: Combine the terms with the same variable, x, in the constraint to simplify it. We get 3x + h < 140.
5. Solve the constraint equation for a single variable: In order to proceed, we will express h in terms of x. Rearrange the equation to isolate h: h < 140 - 3x.
6. Substitute the expression for h into the volume equation: Based on the previous step, we found h to be less than 140 - 3x. Substituting this into the volume equation, it becomes V = 2x^2 * (140 - 3x).
7. Simplify the volume equation: Expand the equation to obtain a quadratic equation in terms of x: V = 280x^2 - 6x^3.
8. Differentiate the volume equation: To find the maximum value of V, we take the derivative of the volume equation with respect to x. The derivative of V = 280x^2 - 6x^3 is dV/dx = 560x - 18x^2.
9. Set the derivative equal to zero and solve for x: To find critical points, set dV/dx = 560x - 18x^2 = 0. Factor out common terms: x(560 - 18x) = 0. Thus, x = 0 or x = 560/18.
10. Determine the valid value for x: Since the length of a side cannot be negative, the only solution is x = 560/18.
11. Substitute the value of x into the constraint equation: Use x = 560/18 in the constraint equation, 3x + h < 140. Solve for h: 3(560/18) + h < 140. Simplify: 560/6 + h < 140. Further simplify: 560/6 + h < 140(6/6). Ultimately, h < 140 - 560/6.
12. Calculate the maximum volume: Substitute the found values of x and h into the volume equation, V = 2(560/18)^2 * [140 - 560/6]. Evaluate this expression to obtain the maximum volume.
By following these steps, you will be able to determine the dimensions of the box with maximum volume, given the constraints provided.