91Ó°ÊÓ

In many optimization problems, calculating the volume is a crucial step. For a box with a square base, the volume can be found using the formula: \ \ \[ V = x^2h \] Here, \(x\) represents the side length of the base, and \(h\) denotes the height of the box. If given a fixed volume, you can solve for one dimension in terms of the other. For example, if the volume \(V\) is 12 cubic feet: \ \ \[ x^2h = 12 \] Solving for \(h\) gives: \ \ \[ h = \frac{12}{x^2} \] Understanding how to express \(h\) in terms of \(x\) will be essential for further calculations in problems involving surface area or cost.

The surface area of a closed rectangular box with a square base includes the areas of the top, bottom, and four sides. For a box with side length \(x\) and height \(h\), the areas can be broken down as follows:

• Top and Bottom: Each \(x^2\)
• Four Sides: Each \(xh\)

Combining these, the total surface area is: \[ A = 2x^2 + 4xh \] By using the earlier expression for \(h\), we substitute and get: \[ 4x \cdot \frac{12}{x^2} = \frac{48}{x} \] So, the surface area expressed in terms of \(x\) is: \[ A = 2x^2 + \frac{48}{x} \] These calculations simplify the cost function, which we'll need to minimize.

In problems like constructing a box, minimizing cost is typically the goal. Costs depend on material types and their respective surface areas. Let's say the top is made of expensive metal at \(\$2\) per square foot, and the rest of the box is made of wood at \(\$1\) per square foot. The cost function can be expressed as a sum of the surface area costs: For top (metal): \ \ \[ \text{Cost}_{\text{metal}} = 2 \cdot x^2 \] For sides and bottom (wood): \ \ \[ \text{Cost}_{\text{wood}} = x^2 + \frac{48}{x} \] Therefore, the total cost function becomes: \ \ \[ \text{Cost} = 2x^2 + x^2 + \frac{48}{x} = 3x^2 + \frac{48}{x} \] Minimizing this cost function requires finding its critical points using derivatives.

Derivatives help us find the critical points of a function, which are potential minima or maxima. The first derivative of a function \(C(x)\) gives the rate of change of the cost with respect to \(x\). The critical points occur when \(C'(x) = 0\). Let's start with the cost function: \ \ \[ C(x) = 3x^2 + \frac{48}{x} \] Its first derivative is: \ \ \[ C'(x) = 6x - \frac{48}{x^2} \] To find the critical points: \ \ \[ 6x - \frac{48}{x^2} = 0 \] Solving for \(x\) yields: \ \ \[ 6x^3 = 48 \Rightarrow x^3 = 8 \Rightarrow x = 2 \] This critical point will help identify if it is the minimum cost. Next, we apply the second derivative test.

The second derivative test determines the concavity of the function at critical points. If \(C''(x) > 0\), the function has a local minimum; if \(C''(x) < 0\), it has a local maximum. Starting with: \[ C(x) = 3x^2 + \frac{48}{x} \] The second derivative is: \[ C''(x) = 6 + \frac{96}{x^3} \] Evaluating at \(x = 2\): \[ C''(2) = 6 + \frac{96}{8} = 6 + 12 = 18 \] Since \(C''(2) > 0\), \(x = 2\) yields a local minimum. This confirms that the dimensions that minimize the cost are:

• Side length of base: \(2\) feet
• Height: \(3\) feet (since \(h = \frac{12}{2^2}\))

Understanding derivatives and the second derivative test is essential for solving cost minimization problems.