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Chapter 2: Problem 6

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 (b) and \((c) ?\)

Short Answer

Expert verified
2) What is the optimal value of t that maximizes the volume of the tray? 3) Does the volume of oil that the tray can carry have a maximum value? 4) For the constrained maximum problem, approximately what are the values of x and t that maximize the volume of the tray, given the constraint of 1,000,000 square feet of steel sheet? Answers: 1) The final expression for the volume of the tray, represented by V, is V = 3tx^2 - 8t^2x + 4t^3. 2) The optimal value of t that maximizes the volume of tray is t = 2x. 3) The volume of oil that the tray can carry does not have a maximum value; it increases indefinitely as x increases. 4) For the constrained maximum problem, the values of x and t that maximize the volume of the tray are approximately x ≈ 119.51 and t ≈ 239.02.

Step by step solution

01

(Step 1: Express the Volume Function)

We start by expressing the volume of the tray in terms of the given parameters. The volume of a rectangular prism, which is what the tray is, can be given by the product of its length, width, and height. In this case, when a square of side \(t\) is cut from each corner, the dimensions of the tray become \((3x-2t)\) for the length, \((x-2t)\) for the width and \(t\) for the height. The volume, \(V\), is therefore given by the following expression: \\[ V = t(3x - 2t)(x-2t) \\] Now we will simplify this expression.
02

(Step 2: Simplify the Volume Expression)

To simplify the volume expression, we will distribute the \(t\) to the other terms and then expand the product: \\[ V = 3tx^2 - 8t^2x + 4t^3 \\] This is the expression we were asked to show in part (a) of the exercise.
03

(Step 3: Maximize the Volume for any Given Value of x)

To find the value of \(t\) that maximizes the volume \(V\) for any given value of \(x\), we take the derivative of the volume expression with respect to \(t\). By setting the derivative equal to 0, we find the critical points of the function: \\[ \frac{dV}{dt} = 4t^2 - 16xt + 12x^2 \\] Setting this expression to 0 and solving for \(t\), we can derive a formula for the optimal value of \(t\): \\[ 4t^2 - 16xt + 12x^2=0 \\] This quadratic equation can be factored as follows: \\[ (2t - 4x)(2t - 3x)=0 \\] Therefore, we get two possible values for \(t\): \\[ t = 2x \qquad or \qquad t = \frac{3}{2}x \\] Because \(t\) represents the length of the squares cut from each corner and we have a constraint that \(t < \frac{1}{2}x\) (the largest possible square that could be cut), we see that the second option \(t = \frac{3}{2}x\) is not possible. Therefore, the value of \(t\) that maximizes \(V\) is: \\[ t = 2x \\]
04

(Step 4: Determine if a Value of x Maximizes the Volume of the Tanker)

Now we will investigate if there is a value of \(x\) that maximizes the volume of oil that can be carried. We already found the optimal value for \(t\): \(t = 2x\). We will now substitute this result back into the volume expression: \\[ V = 3(2x)x^2 - 8(2x^2)x + 4(2x^3) \\] Slickly the expression: \\[ V = 12x^3 \\] The only way for this expression to have a maximum is if its derivative changes from positive to negative. Taking the derivative with respect to \(x\), we get: \\[ \frac{dV}{dx} = 36x^2 \\] The second derivative, in this case, is \(\frac{d^2V}{dx^2} = 72x\), which is positive for all positive values of \(x\). This means there is no value of \(x\) that maximizes the volume of oil that can be carried; it will increase indefinitely as \(x\) increases.
05

(Step 5: Constrained Maximum Problem)

Now we will consider the case where the shipbuilder is constrained to use only 1,000,000 square feet of steel sheet. This constraint is expressed as: \\[ 3x^2-4t^2=1,000,000 \\] We also have the relation \(t = 2x\). Replacing \(t\) in the above equation, we get: \\[ 3x^2 - 4(2x)^2 = 1,000,000 \\] Simplifying the expression: \\[ 7x^2=1,000,000 \\] Solving for \(x\): \\[ x = \sqrt{\frac{1,000,000}{7}} \approx 119.51 \\] With this value of \(x\), the corresponding value of \(t\) can be found: \\[ t = 2x \\] Substituting the value of \(x\): \\[ t \approx 239.02 \\] The solution in this constrained maximum problem is \(x \approx 119.51\) and \(t \approx 239.02\), whereas in parts (b) and (c), we found that \(t = 2x\) and there is no value of \(x\) that maximizes the volume of the oil that can be carried.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Volume Maximization
Understanding the concept of volume maximization is crucial when dealing with physical structures like oil tankers. The goal is to determine the dimensions that will allow the maximum amount of oil to be carried. In microeconomic theory, this sort of problem is equivalent to maximizing the profit or output of a product within given constraints. Here, the problem starts by establishing a relationship between the volume of an oil tanker, represented by a traylike structure, and the dimensions of the steel sheet used to construct it.

As identified in the exercise, the steel sheet starts as a rectangle and squares of side t are cut from each of its corners. Folding up the resulting flaps forms the tray. The key is to express the volume as a function of t, which is what the formula V = 3tx^2 - 8t^2x + 4t^3 represents. When trying to maximize this volume for a given x (width of the original steel sheet), it involves calculating the value of t that results in the largest possible volume. This approach not only helps in optimizing the design of oil tankers but also enhances understanding of the principles behind volume maximization in more complex systems.
Constrained Optimization
The method of constrained optimization is commonly used in various fields, including economics and engineering, to find the best possible solution to a problem within certain limitations. A constraint could be a budget, available resources, physical limitations, or any other factor that restricts the set of possible solutions. In our exercise, the shipbuilder must maximize the volume of the oil tanker without exceeding the 1,000,000 square feet limit of steel sheet.

This real-world constraint shapes the optimization process and creates a more challenging problem, as compared to the scenario where there are no upper bounds on the resources. The provided equation, 3x^2 - 4t^2 = 1,000,000, is the mathematical representation of the constraint and plays a critical role in finding the optimal dimensions of the tanker. Techniques in constrained optimization allow the balancing of an objective against limitations, ensuring an optimal solution that still meets all necessary requirements.
Quadratic Equations
At the heart of this optimization problem lies a quadratic equation, which is a second-degree polynomial equation of the form ax^2 + bx + c = 0. Quadratic equations are omnipresent in various mathematical problems and can describe a wide range of phenomena from projectile motion to the shape of satellite dishes.

In the context of the oil tanker volume maximization problem, the quadratic equation arises when setting the first derivative of the volume function to zero, which is a standard method to find the maximum or minimum point of a function. Solving the quadratic equation 4t^2 - 16xt + 12x^2 = 0 provides possible values for t that could potentially maximize the volume. However, since t must be less than half of x, this narrows down the options. Understanding how to manipulate and solve quadratic equations is vital for students as it equips them to handle a broad spectrum of problems in both academic and real-world applications.

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Most popular questions from this chapter

Consider the following constrained maximization problem: \\[ \begin{array}{ll} \text { maximize } & y=x_{1}+5 \ln x_{2} \\ \text { subject to } & k-x_{1}-x_{2}=0 \end{array} \\] where \(k\) is a constant that can be assigned any specific value. a show that if \(k=10,\) this problem can be solved as one involving only equality constraints. b. Show that solving this problem for \(k=4\) requires that \\[ x_{1}=-1 \\] c. If the \(x\) 's in this problem must be non-negative, what is the optimal solution when \(k=4 ?\) (This problem may be solved cither intuitively or using the methods outlined in the chapter.) d. What is the solution for this problem when \(k=20 ?\) What do you conclude by comparing this solution with the solution for part (a)?

Suppose a firm's total revenues depend on the amount produced \((q)\) according to the function \\[ R=70 q-q^{2} \\] Total costs also depend on \(q\) \\[ C=q^{2}+30 q+100 \\] a. What level of output should the firm produce to maximize profits \((R-C) ?\) What will profits be? b. Show that the second-order conditions for a maximum are satisfied at the output level found in part (a). c. Does the solution calculated here obey the "marginal revenue equals marginal cost" rule? Explain.

Taylor's theorem shows that any function can be approximated in the vicinity of any convenient point by a series of terms involving the function and its derivatives, Here we look at some applications of the theorem for functions of one and two variables. a Any continuous and differentiable function of a single variable, \(f(x),\) can be approximated near the point \(a\) by the formula \\[ \begin{aligned} f(x)=& f(a)+f(a)(x-a)+0.5 f^{\prime \prime}(a)(x-a)^{2}+\\\ &+\text { terms in } f^{\prime \prime \prime}, f^{\prime \prime \prime}, \ldots . \end{aligned} \\] Using only the first three of these terms results in a quadratic Taylor approximation. Use this approximation together with the definition of concavity to show that any concave function must lie on or below the tangent to the function at point \(a\) b. The quadratic Taylor approximation for any function of two variables, \(f(x, y),\) near the point \((a, b)\) is given by \\[ \begin{aligned} f(x, y)=& f(a, b)+f_{1}(a, b)(x-a)+f_{2}(a, b)(y-b) \\ &+0.5\left[f_{11}(a, b)(x-a)^{2}\right.\\\ &\left.+2 f_{12}(a, b)(x-a)(y-b)+f_{22}(y-b)^{2}\right] \end{aligned} \\] Use this approximation to show that any concave function (as defined by Equation 2.84 ) must lie on or below its tangent plane at \((a, b)\)

The height of a ball that is thrown straight up with a certain force is a function of the time ( \(t\) ) from which it is released given by \(f(t)=-0.5 g t^{2}+40 t\) (where \(g\) is a constant determined by gravity). a. How does the value of \(t\) at which the height of the ball is at a maximum depend on the parameter \(g\) ? b. Use your answer to part (a) to describe how maximum height changes as the parameter \(g\) changes. c. Use the envelope theorem to answer part (b) directly. d. On the Earth \(g=32,\) but this value varies somewhat around the globe. If two locations had gravitational constants that differed by \(0.1,\) what would be the difference in the maximum height of a ball tossed in the two places?

Because the expected value concept plays an important role in many economic theories, it may be useful to summarize a few more properties of this statistical measure. Throughout this problem, \(x\) is assumed to be a continuous random variable with PDF \(f(x)\) a. (Jensen's inequality) Suppose that \(g(x)\) is a concave function. Show that \(E[g(x)] \leq g[E(x)] .\) Hint: Construct the tangent to \(g(x)\) at the point \(E(x)\). This tangent will have the form \(c+d x \geq g(x)\) for all values of \(x\) and \(c+d E(x)=g[E(x)],\) where \(c\) and \(d\) are constants. b. Use the procedure from part (a) to show that if \(g(x)\) is a convex function, then \(E[g(x)] \geq g[E(x)]\) c. Suppose \(x\) takes on only non-negative values-that is, \(0 \leq x \leq \infty\). Use integration by parts to show that \\[ E(x)=\int[1-F(x)] d x \\] where \(F(x)\) is the cumulative distribution function for \(x\) \\[ \left[\text { ice, } F(x)=\int_{0}^{x} f(t) d t\right] \\] d. (Markov's inequality) Show that if \(x\) takes on only positive values, then the following inequality holds: \\[ \begin{array}{c} P(x \geq t) \leq \frac{E(x)}{t} \\ \text { Hint: } E(x)=\int_{0}^{\infty} x f(x) d x=\int_{0}^{t} x f(x) d x+\int_{t}^{\infty} x f(x) d x \end{array} \\] e. Consider the PDF \(f(x)=2 x^{-3}\) for \(x \geq 1\) 1\. Show that this is a proper PDF. 2\. Calculate \(F(x)\) for this PDF. 3\. Use the results of part (c) to calculate \(E(x)\) for this \(P D F\) 4\. Show that Markov's incquality holds for this function. f. The concept of conditional expected value is useful in some economic problems. We denote the expected value of \(x\) conditional on the occurrence of some event, \(A\), as \(E(x | A),\) To compute this value we need to know the PDF for \(x \text { given that } A \text { has occurred [denoted by } f(x | A)]\) With this notation, \(E(x | A)=\int_{-\infty}^{+\infty} x f(x | A) d x .\) Perhaps the easiest way to understand these relationships is with an example. Let \\[ f(x)=\frac{x^{2}}{3} \quad \text { for } \quad-1 \leq x \leq 2 \\] 1\. Show that this is a proper PDF. 2\. Calculate \(E(x)\) 3\. Calculate the probability that \(-1 \leq x \leq 0\) 4\. Consider the event \(0 \leq x \leq 2,\) and call this event \(A\). What is \(f(x | A) ?\) 5\. Calculate \(E(x | A)\) 6\. Explain your results intuitively.
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