Problem 16 - Active Calculus Multivariable

Question

Design a rectangular milk carton box of width \(w\), length \(l\), and height \(h\) which holds \(500 \mathrm{~cm}^{3}\) of milk. The sides of the box cost 1 cent \(/ \mathrm{cm}^{2}\) and the top and bottom cost \(2 \mathrm{cent} / \mathrm{cm}^{2}\). Find the dimensions of the box that minimize the total cost of materials used. dimensions = __________.

Step-by-Step Solution

Answer

dimensions = \(\sqrt[3]{\frac{500}{2}} \times \sqrt[3]{\frac{500}{4}} \times \sqrt[3]{\frac{1000}{4}}\).

Step 1: Write down the cost equation

Considering the dimensions \(w, l, h,\) the respective areas of the sides, top, and bottom are: \(wh\), \(lh\), and \(2lw\). Since the sides of the box cost \(1\) cent per cm², and the top and bottom cost \(2\) cents per cm², the cost function \(C(w, l, h)\) can be written as: \[C(w, l, h) = (2wh + 2lh) \cdot 1 + 2lw \cdot 2\] or, simplifying, \[C(w, l, h) = 2wh + 2lh + 4lw\] This is the total cost of materials used for the box.

Step 2: Write down the volume constraint

The volume \(V\) of the milk carton box is given by the product of its dimensions, that is: \[V = w \cdot l \cdot h\] Since the box must hold \(500\mathrm{~cm}^3\) of milk, we have: \[500 = w \cdot l \cdot h\]

Step 3: Express \(h\) in terms of \(V, w,\) and \(l\)

We can rewrite the volume constraint equation as: \[h = \frac{500}{w \cdot l}\]

Step 4: Substitute the expression of \(h\) in the cost function

By using the expression for \(h\) found in Step 3, we can substitute it into the cost function: \[C(w, l, h) = 2w \left(\frac{500}{wl}\right) + 2l \left(\frac{500}{wl}\right) + 4lw\] Simplify the cost function: \[C(w, l) = \frac{1000}{l} + \frac{1000}{w} + 4lw\] The cost function now only depends on two variables, \(w\) and \(l\).

Step 5: Find critical points of the cost function

To find the critical points of the cost function, we'll take the partial derivatives of \(C(w, l)\) with respect to \(w\) and \(l\), and set them equal to zero: 1. \(\frac{\partial C}{\partial w} = -\frac{1000}{w^2} + 4l = 0\) 2. \(\frac{\partial C}{\partial l} = -\frac{1000}{l^2} + 4w = 0\)

Step 6: Solve equations to find the dimensions

We will now solve these two equations simultaneously. First, solve equation (1) for \(l\), noting that \(l=\frac{1000}{4w^2}\). Now, substitute this expression of \(l\) into equation (2): \[-\frac{1000}{\left(\frac{1000}{4w^2}\right)^2} + 4w = 0\] Solving this equation for \(w\), we get: \[w = \sqrt[3]{\frac{500}{2}}\] Now, by substituting this \(w\) back into the expression for \(l=\frac{1000}{4w^2}\), we get: \[l = \sqrt[3]{\frac{500}{4}}\] Finally, using the expression for \(h\) from Step 3, we get: \[h = \frac{500}{w \cdot l} = \sqrt[3]{\frac{1000}{4}}\] dimensions = \(\sqrt[3]{\frac{500}{2}} \times \sqrt[3]{\frac{500}{4}} \times \sqrt[3]{\frac{1000}{4}}\).

Key Concepts

Cost Function

The concept of a cost function is central in various optimization problems. It's an equation designed to estimate the total cost of producing or constructing an item, by considering different contributing factors. In the context of our milk carton box problem, the cost function is based on how much it costs to cover the box with material. The sides of the box cost 1 cent per square centimeter, while the top and bottom cost 2 cents per square centimeter.
This leads us to the cost function:

• Each side of the box contributes a cost based on its area, multiplied by its cost rate. For instance, the areas of the sides are computed using the height (h) and width (w) or length (l), i.e., 2wh and 2lh.
• The top and bottom surfaces, with area 2lw, contribute to the cost at a higher rate due to their pricing.

Hence, the overall cost function that needs to be minimized is: \[ C(w, l, h) = 2wh + 2lh + 4lw \]The goal here is to minimize this cost by adjusting the dimensions of the box.

Volume Constraint

Volume constraint refers to a fixed requirement that the object in question must adhere to in terms of its volume. In our case, the milk carton must hold exactly 500 cm³ of milk, which becomes a non-negotiable constraint in our optimization problem.
Given the box's dimensions, its volume V is simply the product of its width (w), length (l), and height (h):

• This gives us the equation \[ V = w \times l \times h \]

Since the volume is fixed at 500 cm³, this relationship becomes:\[ 500 = w \times l \times h \]This equation allows us to express one variable, say height, in terms of the other two variables (width and length), simplifying the problem. Eventually, this constraint tightly binds our dimensions, steering the optimization process.

Partial Derivatives

Partial derivatives are powerful tools in calculus to explore how a function's output changes as one specific variable changes, while the others are held constant. In optimization problems, they are critical in finding the minimum or maximum values of a function.
For the milk carton problem, we're interested in how changes in width (w) and length (l) affect the total cost. We compute the partial derivatives of the cost function:

• With respect to the width, \( \frac{\partial C}{\partial w} \) gives insight into how cost changes if width varies.
• Similarly, \( \frac{\partial C}{\partial l} \) assesses changes due to variations in length.

Through these derivatives, we identify where these changes become negligible, leading us to possible critical points. Setting these partial derivatives to zero helps determine conditions for minimal cost.

Critical Points

Critical points are key in understanding where a function might reach a local minimum or maximum. In the world of optimization, our aim is to locate these points to find the most cost-effective design for our milk carton.
In practice, a critical point is where the derivative of a function equals zero. Here, for the milk carton, it involves solving:

• \( \frac{\partial C}{\partial w} = 0 \), and
• \( \frac{\partial C}{\partial l} = 0 \)

Solving these equations together allows us to find values for w and l, where changes in either direction cause the least increase in the cost. These are the dimensions that minimize our construction expenditures. By checking second derivative tests or using algebraic solutions, these computations confirm the optimal size for minimal material cost.