Question

A simple way to model the construction of an oil tanker is to start with a large rectangular sheet of steel that is \(x\) feet wide and \(3 x\) feet long. Now cut a smaller square that is \(t\) feet on a side out of each corner of the larger sheet and fold up and weld the sides of the steel sheet to make a traylike structure with no top. a. Show that the volume of oil that can be held by this tray is given by \\[ V=t(x-2 t)(3 x-2 t)=3 t x^{2}-8 t^{2} x+4 t^{3} \\] b. How should \(t\) be chosen to maximize \(V\) for any given value of \(x ?\) c. Is there a value of \(x\) that maximizes the volume of oil that can be carried? d. Suppose that a shipbuilder is constrained to use only 1,000,000 square feet of steel sheet to construct an oil tanker. This constraint can be represented by the equation \(3 x^{2}-4 t^{2}=1,000,000\) (because the builder can return the cut-out squares for credit). How does the solution to this constrained maximum problem compare with the solutions described in parts and (c)?

Short answer

Short Answer: To maximize the volume of the oil held by the tray, we first found the equation for the volume V as V = 3tx^2 - 8t^2x + 4t^3 and then optimized V in terms of t. The first derivative with respect to t is dV/dt = 3x^2 - 16tx + 12t^2, which needs to be set to zero to find the optimal value of t. The second derivative, d^2V/dt^2, depends both on t and x, so we cannot find a maximum value solely for x. Finally, under the constraint 3x^2 - 4t^2 = 1,000,000, solving the constrained maximum problem will yield a different result than the solutions found in parts (b) and (c).

Step-by-step solution

Understand the Geometry and Dimensions

From the description, we have a large rectangular sheet with width x feet and length 3x feet, and we cut out t feet squares from each corner, folding along the sides. The resulting tray-like structure will have a width of (x - 2t), a length of (3x - 2t), and a height of t.

Write the Volume Equation

The volume V of the tray-shaped oil tanker can be found by multiplying the width, length, and height: V = t(x - 2t)(3x - 2t) Now we can expand and simplify the equation to get: V = 3tx^2 - 8t^2x + 4t^3 This is the equation we want to analyze.

Maximize V with respect to t

Since we want to maximize V in terms of t, we can find its first derivative with respect to t, and set it equal to 0. dV/dt = d(3tx^2 - 8t^2x + 4t^3)/dt = 3x^2 - 16tx + 12t^2 Now set the above equation equals to zero: 3x^2 - 16tx + 12t^2 = 0 We can use this equation to find the optimal value of t for any given x to maximize the volume of our tray.

Analyze if there's a maximum value for x

To determine if there's a maximum value for x that leads to the maximum volume, we need to study the second derivative of our volume equation V(t, x). By doing so, we can find any points of concavity to determine if a maximum value exists. Start by finding the second derivative d^2V/dt^2: d^2V/dt^2 = d(3x^2 - 16tx + 12t^2)/dt = -16x + 24t Since the second derivative does not depend only on x, we cannot directly find a maximum value for x in this problem.

Solve the constrained maximum problem

Under the constraint given by the equation 3x^2 - 4t^2 = 1,000,000, we can solve the constrained maximum problem and compare it to the solution found in parts (b) and (c). We can isolate t^2 from the constraint equation: t^2 = (3x^2 - 1,000,000) / 4 Now substitute this expression for t^2 into the equation that we derived in Step 3. 3x^2 - 16tx + 3x^2 - 1,000,000 = 0 We'll now need to solve the above equation for x, and then find the corresponding value of t using the constraint equation. Overall, the solution to this constrained maximum problem will yield a different result than those described in parts (b) and (c).

Key concepts

Volume Maximization

When working with constrained optimization problems such as maximizing the volume of an oil tanker, understanding the volume function is crucial. In this exercise, we derive the volume expression for a tray-like structure that will serve as the oil tanker. The tanker's volume can be represented through the function:\[V = t(x - 2t)(3x - 2t) = 3tx^2 - 8t^2x + 4t^3\]The formula arises from multiplying the dimensions:- Width: \(x - 2t\)- Length: \(3x - 2t\)- Height: \(t\)These dimensions stem from cutting squares from each corner of a rectangular steel sheet and folding the sides. Thus, the expression \(V = 3tx^2 - 8t^2x + 4t^3\) helps determine how to adjust \(t\), the side of the squares, to achieve maximum volume. This part of the problem sets the baseline for understanding volume maximization using a practical geometric context.

Derivative Analysis

Analyzing derivatives is essential to optimization problems as they help identify points that maximize or minimize a given function. For this volume maximization task, we find the first derivative of the volume function, \(V\), with respect to \(t\):\[\frac{dV}{dt} = 3x^2 - 16tx + 12t^2\]To find critical points where volume is maximized, set \(\frac{dV}{dt} = 0\):\[3x^2 - 16tx + 12t^2 = 0\]This equation allows us to solve for \(t\) in terms of \(x\), identifying the optimal cut-out size for each possible dimension of the steel sheet.

Additionally, the second derivative, \(\frac{d^2V}{dt^2} = -16x + 24t\), helps confirm if these points are maxima. If \(\frac{d^2V}{dt^2}\) is negative at a critical point, it suggests a local maximum in volume.

The derivative analysis provides insight into how various values of \(t\) and \(x\) affect the volume and aids in ensuring that volume is maximized effectively.

Oil Tanker Construction

In oil tanker construction, using constrained resources efficiently is key. Here, the problem presents a scenario where only 1,000,000 square feet of steel is available.

This leads to the constraint equation:\[3x^2 - 4t^2 = 1,000,000\]The shipbuilder can use this relation to figure out suitable values of \(t\) and \(x\) that conform to the material limits while maximizing volume. When you solve the constraint, you determine a feasible range for these dimensions.

Re-examining the formula derived for \(t^2\):\[t^2 = \frac{3x^2 - 1,000,000}{4}\]This expression substitutes back into the derivative equation, showing how the constrained solution varies from the unconstrained one. Balancing the material constraints with optimization goals is critical in practical construction scenarios.

This exercise illustrates essential principles in optimizing designs under specific limitations, a common theme in engineering fields, especially in large-scale projects like oil tanker construction.