Problem 8 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

Find the largest box that will fit in the positive octant \((x \geq 0, y \geq 0,\) and \(z \geq 0)\) and underneath the plane \(z=12-2 x-3 y\).

Step-by-Step Solution

Answer

The largest box has a volume of 16 cubic units.

Step 1: Understand the Problem

We need to find the largest rectangular box that fits beneath the plane \(z = 12 - 2x - 3y\) and is within the positive octant \(x \geq 0, y \geq 0, z \geq 0\). The box has vertices at the origin \((0,0,0)\) and \((x,y,z)\) where \(z\) follows the plane's constraint.

Step 2: Express the Volume of the Box

The volume \(V\) of the box can be expressed as a function of \(x\) and \(y\), where the height \(z = 12 - 2x - 3y\):\[V = x \cdot y \cdot z = x \cdot y \cdot (12 - 2x - 3y)\]

Step 3: Simplify the Volume Function

Substitute the expression for \(z\) into the equation to get the volume of the box:\[V = x \cdot y \cdot (12 - 2x - 3y) = 12xy - 2x^2y - 3xy^2\]

Step 4: Find Critical Points

To find the maximum volume, take partial derivatives with respect to \(x\) and \(y\), set them to zero, and solve the resulting system:\[\frac{\partial V}{\partial x} = 12y - 4xy - 3y^2 = 0\]\[\frac{\partial V}{\partial y} = 12x - 2x^2 - 6xy = 0\]

Step 5: Solve the System of Equations

Solve the system:\[12y - 4xy - 3y^2 = 0\] Factor: \( y(12 - 4x - 3y) = 0 \) giving \(y = 0\) or \(12 - 4x - 3y = 0\).For \(12 - 4x - 3y = 0\), substitute \(y\) into the second equation: \(12x - 2x^2 - 6x(\frac{12 - 4x}{3}) = 0\). Solve stepwise to find critical points.

Step 6: Check Feasibility and Constraints

Solve and confirm the feasibility: First critical point from solving gives \(x = 2, y = 2\) with \(z = 12 - 2(2) - 3(2) = 4\). The vertex \((2, 2, 4)\) lies in the positive octant, and under the plane.

Step 7: Compute the Maximum Volume

Substitute \(x = 2, y = 2\) into the volume equation: \(V = 2 \times 2 \times 4 = 16\). Thus, the maximum volume possible for the box is 16 cubic units.

Key Concepts

Partial Derivatives

Partial derivatives are a core part of calculus optimization. They help us analyze functions with more than one variable. In the context of this exercise, we have a function of two variables:

• The volume function: \[ V = 12xy - 2x^2y - 3xy^2 \]

We are interested in how changes in one variable (like \(x\)) impact the function while keeping the other variable (\(y\)) constant, and vice versa. This is where partial derivatives come in.

• The partial derivative with respect to \(x\) is derived by treating \(y\) as a constant:\[ \frac{\partial V}{\partial x} = 12y - 4xy - 3y^2 \]
• Similarly, the partial derivative with respect to \(y\) treats \(x\) as a constant:\[ \frac{\partial V}{\partial y} = 12x - 2x^2 - 6xy \]

To find optimal values for \(x\) and \(y\) that maximize the volume, we set these partial derivatives to zero, leading us to critical points.

Critical Points

Critical points occur where the partial derivatives of a function are zero. These points are potential sites for optimization, such as finding maximum or minimum values.
In the exercise, after computing the partial derivatives and setting them to zero:

• \[ \frac{\partial V}{\partial x} = 12y - 4xy - 3y^2 = 0 \]
• \[ \frac{\partial V}{\partial y} = 12x - 2x^2 - 6xy = 0 \]

We solve these simultaneous equations to identify critical points. In this example, these critical point equations suggest locations where changes in dimensions result in no further increase in volume.
Solving gives us critical points, such as \( (x, y) = (2, 2) \). These values are inputs for testing whether they provide the maximum volume any other condition violates or doesn't satisfy.

Volume Maximization

Volume maximization involves computing the shape's highest possible volume within set constraints. In this problem, the constraints include remaining below the plane and within the positive octant.
Using the critical point found (\(x = 2, y = 2\)), we substitute back into the volume function to confirm if it truly maximizes the volume:

• Substituting places the vertex at \((2, 2, z)\) with the height:\[ z = 12 - 2(2) - 3(2) = 4 \]
• Calculating the volume:\[ V = 2 \times 2 \times 4 = 16 \]

This gives a maximum volume of 16 cubic units while ensuring the conditions of the exercise are met. The combination of partial derivatives and critical points helps guide us to this optimal configuration.