Question

A lidless box is to be made using \(2 \mathrm{m}^{2}\) of cardboard. Find the dimensions of the box with the largest possible volume.

Short answer

Answer: To maximize the volume of a lidless box made of 2 square meters of cardboard, the dimensions should be approximately x = ___ m, y = ___ m, and z = ___ m.

Step-by-step solution

Define the variables and set up an equation for the volume

Let's denote the length of the base as 'x', the width as 'y', and the height as 'z'. The volume of the box V can be expressed as the product of its dimensions: V = x*y*z

Create an equation relating the cardboard area constraint to the dimensions of the box

The total area of cardboard used to construct this lidless box is \(2 \mathrm{m}^{2}\). Since there is no lid, we are only using the bottom, two sides, and two ends of the box to form the area constraint. Let's write an equation representing the total area using x, y, and z: 2 = xy + 2(xz) + 2(yz)

Solve the area constraint equation for one variable

We can solve this equation for one of the variables. Let's solve it for z: z = (2 - xy) / (2x + 2y)

Substitute the value of z into the volume equation

Now that we have the expression for z in terms of x and y, substitute this equation into the original volume equation: V(x,y) = x * y * ( (2 - xy) / (2x + 2y) )

Differentiate the volume equation with respect to x and y to find the critical points

In order to maximize the volume, we need to find the critical points by taking the partial derivative of the volume with respect to x and y and setting these derivatives to zero: V_x = ∂V/∂x V_y = ∂V/∂y Now set V_x = 0 and V_y = 0 to find the critical points.

Find the critical points for the volume equation by solving V_x = 0 and V_y = 0

After solving V_x = 0 and V_y = 0 simultaneously, we will get the critical points (x,y) that will help us to determine the dimensions of the box that will maximize its volume.

Plug the critical point values back into the equation relating dimensions to find the value of z

Based on the critical point values, substitute x and y back into the equation relating the dimensions and solve for the height z.

Verify the maximum volume by checking the second-order partial derivatives

It's important to verify that the volume is actually maximized and not minimized at the critical points. Compute the second-order partial derivatives V_xx, V_yy, and V_xy and use these values to find the Hessian determinant. If the determinant is positive and V_xx is negative, the volume is maximized.

Report the dimensions of the box that maximize its volume

After completing the steps above, we will know the dimensions x, y, and z that maximize the volume of the lidless box made of \(2 \mathrm{m}^{2}\) of cardboard.

Key concepts

Volume Maximization

Maximizing the volume of a lidless box is an optimization problem where we use given constraints to determine the dimensions that result in the largest possible volume. Imagine using a fixed amount of cardboard, specifically \(2 \, \text{m}^2\), to construct the box.

The volume \(V\) of a box is calculated by multiplying its length, width, and height, which for a box with dimensions \(x\), \(y\), and \(z\) can be expressed as:
\[ V = x \cdot y \cdot z \]
To find the optimal dimensions, we need to account for the area constraint due to the limitation of the cardboard, ensuring we never exceed \(2 \, \text{m}^2\). This forms the basis of finding a solution that maximizes the volume with the given material constraints. Solving optimization problems with constraints often involves using calculus techniques, particularly when it comes to non-linear relationships between variables.

Critical Points

In mathematical optimization, critical points are values of variables where a function's derivative is zero, signaling potential maximum or minimum values. To solve for a lidless box's maximum volume, we find the critical points from the volume's partial derivatives with respect to the variables \(x\) and \(y\).

Finding critical points involves:

• Calculating the partial derivatives: \(V_x = \frac{\partial V}{\partial x}\) and \(V_y = \frac{\partial V}{\partial y}\).
• Setting these partial derivatives equal to zero to obtain equations: \(V_x = 0\) and \(V_y = 0\).
• Solving these simultaneously to get the values of \(x\) and \(y\) that make the derivatives zero.

The solutions yield the points at which the volume may achieve either a local maximum or minimum. These points, by themselves, might not give enough insight, so further analysis is crucial to discern if they lead to a volume maximization.

Partial Derivatives

Partial derivatives help us understand how a multi-variable function's value changes concerning changes in each variable, one at a time. For functions of several variables, computing partial derivatives is a critical tool in optimization.

When dealing with the volume equation \(V(x, y)\), finding partial derivatives \(\frac{\partial V}{\partial x}\) and \(\frac{\partial V}{\partial y}\) involves differentiation while treating other variables as constants. This indicates how the volume would change by varying \(x\) or \(y\) while holding the other constant.

• Essentially, \(\frac{\partial V}{\partial x}\) shows volume change due to a small change in \(x\).
• Likewise, \(\frac{\partial V}{\partial y}\) reveals the impact of a small change in \(y\).

These derivative values are crucial in setting up the equations for finding the critical points, which aids in determining the conditions for maximizing the box's volume subject to the constraints.

Hessian Determinant

The Hessian determinant is a calculus tool used for analyzing the nature of critical points for functions of multiple variables. It comprises second-order partial derivatives, providing insight into whether a function has a local maximum, minimum, or saddle point at the critical points found.

For a function \(f(x, y)\), the Hessian matrix is:
\[H = \begin{bmatrix}\frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2}\end{bmatrix}\]

The determinant of this matrix evaluates the nature of critical points for the volume function \(V\):

• If the Hessian determinant is positive and \(\frac{\partial^2 V}{\partial x^2} < 0\), then the function has a local maximum at those critical points.
• If it's positive and \(\frac{\partial^2 V}{\partial x^2} > 0\), then it's a local minimum.
• An negative Hessian determinant indicates a saddle point.

This analysis lets us confirm whether the critical points we've identified indeed represent maximum volume for our cardboard box. This application ensures we're solving for the box's optimal dimensions accurately.