Problem 10 - Active Calculus Multivariable

Question

Find the maximum and minimum volumes of a rectangular box whose surface area equals 7000 square \(\mathrm{cm}\) and whose edge length (sum of lengths of all edges) is \(440 \mathrm{~cm}\). Maximum value is ________________. occuring at ( ____________, ______________) Manimum value is ________________. occuring at ( ____________, ______________)

Step-by-Step Solution

Answer

Maximum value is \(\cfrac{121000}{3}\) cm³. Occurring at \(\left(\cfrac{55}{3}(3,11)\right)\). Minimum value is \(\cfrac{605000}{11}\) cm³. Occurring at \(\left(\cfrac{55}{11}(11,3)\right)\).

Step 1: Express the surface area and sum of edge lengths in terms of box dimensions

Let the dimensions of the box be length \(l\), width \(w\), and height \(h\). The surface area (S) of a box is given by the equation \(S = 2lw + 2lh + 2wh\). The sum of edge lengths (E) of the box is given by the equation \(E = 4l + 4w + 4h\). We are given that \(S = 7000\) and \(E = 440\).

Step 2: Derive a relationship between the dimensions

From the equation \(E = 440\), we can write down the expressions for each dimension as follows: \(l = \cfrac{440 - 4w - 4h}{4}\), \(w = \cfrac{440 - 4l - 4h}{4}\), and \(h = \cfrac{440 - 4l - 4w}{4}\).

Step 3: Express the volume in terms of two dimensions

The volume (V) of a rectangular box is given by the equation \(V = lwh\). We will use the expressions for \(l\) and \(w\) in terms of \(h\) to express the volume as a function of two variables: \(V(h,w) = \left(\cfrac{440 - 4w - 4h}{4}\right)w\left(h\right)\).

Step 4: Find the partial derivatives of the volume function

To find the maximum and minimum volumes, we need to find the critical points of the function \(V(h,w)\). For this, we'll need to find its partial derivatives with respect to both variables, \(h\) and \(w\). \(\cfrac{\partial V}{\partial h} = -\cfrac{1}{4}w(440 - 4w - 8h)\) \(\cfrac{\partial V}{\partial w} = \cfrac{1}{4}h(440 - 8w - 4h)\)

Step 5: Find the critical points of the volume function

To find the critical points, set both partial derivatives equal to zero and solve for \(h\) and \(w\). \(-\cfrac{1}{4}w(440 - 4w - 8h) = 0\) \(\cfrac{1}{4}h(440 - 8w - 4h) = 0\) Solving these equations gives us two critical points: \((h_1,w_1) = \cfrac{55}{3}(3,11)\), and \((h_2,w_2) = \cfrac{55}{11}(11,3)\).

Step 6: Determine maximum and minimum volumes

Plug these points back into the volume function to find their corresponding volumes. \(V_1 = V\left(\cfrac{55}{3}(3,11)\right) = \cfrac{121000}{3}\) \(V_2 = V\left(\cfrac{55}{11}(11,3)\right) = \cfrac{605000}{11}\) Since \(\cfrac{121000}{3} > \cfrac{605000}{11}\), the maximum volume occurs at \((h_1, w_1) = \cfrac{55}{3}(3,11)\), and the minimum volume occurs at \((h_2, w_2) = \cfrac{55}{11}(11,3)\). Final answer: Maximum value is \(\cfrac{121000}{3}\) cm³. Occurring at \(\left(\cfrac{55}{3}(3,11)\right)\). Minimum value is \(\cfrac{605000}{11}\) cm³. Occurring at \(\left(\cfrac{55}{11}(11,3)\right)\).

Key Concepts

Critical Points

To understand optimization problems, it's essential to learn about critical points. Critical points are where a mathematical function's derivative equals zero or is undefined. In simple terms, they are the potential points where a function might achieve its maximum or minimum value. In the context of our exercise, we have a function representing the volume of a rectangular box. To find the critical points, we differentiate this function with respect to its variables, which are the dimensions of the box. After finding the derivatives, we set them equal to zero to solve for the critical points. These points indicate where the function might reach a peak (a maximum) or a trough (a minimum). Identifying these critical points is a crucial step toward optimizing functions and solving problems in geometry and calculus.

Surface Area

The surface area of a shape is the total area covered by the surfaces of that shape. For a rectangular box, this includes the areas of all six rectangular faces. Mathematically, the surface area (\(S\) ) of a box with length \(l\), width \(w\), and height \(h\) is given by \(S = 2lw + 2lh + 2wh\). In optimization problems like the one we're tackling, the surface area often acts as a constraint. This means that all other calculations like volume must be done while respecting the fixed surface area available.

Edge Length

The edge length of a box is the sum of the lengths of all the edges. A rectangular box has 12 edges, and if \(l\), \(w\), and \(h\) represent its dimensions, then the total edge length (\(E\)) would be \(E = 4l + 4w + 4h\). For this particular problem, we have an edge length that equals 440 cm. Like surface area, edge length serves as a constraint in the optimization problem. This constraint helps in forming relationships between the different dimensions of the box, allowing us to express the box’s volume in terms of fewer variables and ultimately find the critical points more easily.

Partial Derivatives

Partial derivatives are a fundamental tool in calculus, especially useful when dealing with functions of multiple variables. A partial derivative measures how a function changes as one of its variables is varied while keeping the others constant.In our exercise, we deal with a volume function of two variables: \(h\) and \(w\). To find the critical points, we calculate the partial derivative of the volume with respect to each variable. For example, the partial derivative of the volume with respect to height (\(h\)) determines how changes in the height affect the volume. Similarly, the partial derivative with respect to width (\(w\)) shows how changes in width affect the volume. By setting these partial derivatives to zero, we find critical points that help determine where the volume of the box is maximized or minimized.