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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 and (c)?

Short Answer

Expert verified
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

01

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.
02

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.
03

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.
04

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.
05

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).

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Most popular questions from this chapter

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_{0}^{\infty}[1-F(x)] d x \\] where \(\left.F(x) \text { is the cumulative distribution function for } x \text { [that is, } F(x)=\int_{0}^{x} f(t) d t\right]\) d. (Markov's inequality) Show that if \(x\) takes on only positive values then the following inequality holds: \\[ \begin{array}{r} 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 PDF. 4\. Show that Markov's inequality 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 \(x\) given that \(A\) 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.

Suppose that \(f(x, y)=x y .\) Find the maximum value for \(f\) if \(x\) and \(y\) are constrained to sum to \(1 .\) Solve this problem in two ways: by substitution and by using the Lagrange multiplier method.

Suppose that a firm has a marginal cost function given by \(M C(q)=q+1\) What is this firm's total cost function? Explain why total costs are known only up to a constant of integration, which represents fixed costs. b. As you may know from an earlier economics course, if a firm takes price ( \(p\) ) as given in its decisions then it will produce that output for which \(p=M C(q)\). If the firm follows this profit-maximizing rule, how much will it produce when \(p=15 ?\) Assuming that the firm is just breaking even at this price, what are fixed costs? c. How much will profits for this firm increase if price increases to \(20 ?\) d. Show that, if we continue to assume profit maximization, then this firm's profits can be expressed solely as a function of the price it receives for its output. e. Show that the increase in profits from \(p=15\) to \(p=20\) can be calculated in two ways: (i) directly from the equation derived in part (d); and (ii) by integrating the inverse marginal cost function \(\left[M C^{-1}(p)=p-1\right]\) from \(p=15\) to \(p=20\) Explain this result intuitively using the envelope theorem.

Another function we will encounter often in this book is the power function: \\[ y=x^{\delta} \\] the form \(y=x^{8} / 8\) to ensure that the derivatives have the proper sign). a. Show that this function is concave (and therefore also, by the result of Problem \(2.9,\) quasi-concave). Notice that the \(\delta=1\) is a special case and that the function is "strictly" concave only for \(\delta<1\) b. Show that the multivariate form of the power function \\[ y=f\left(x_{1}, x_{2}\right)=\left(x_{1}\right)^{8}+\left(x_{2}\right)^{8} \\] is also concave (and quasi-concave). Explain why, in this case, the fact that \(f_{12}=f_{21}=0\) makes the determination of concavity especially simple. c. One way to incorporate "scale" effects into the function described in part (b) is to use the monotonic transformation \\[ g\left(x_{1}, x_{2}\right)=y^{y}=\left[\left(x_{1}\right)^{8}+\left(x_{2}\right)^{8}\right]^{y} \\] where \(\gamma\) is a positive constant. Does this transformation preserve the concavity of the function? Is \(g\) quasi-concave?

Here are a few useful relationships related to the covariance of two random variables, \(x_{1}\) and \(x_{2}\) a show that \(\operatorname{Cov}\left(x_{1}, x_{2}\right)=E\left(x_{1} x_{2}\right)-E\left(x_{1}\right) E\left(x_{2}\right) .\) An important implication of this is that if \(\operatorname{Cov}\left(x_{1}, x_{2}\right)=0, E\left(x_{1} x_{2}\right)\) \(=E\left(x_{1}\right) E\left(x_{2}\right) .\) That is, the expected value of a product of two random variables is the product of these variables' expected values. b. Show that \(\operatorname{Var}\left(a x_{1}+b x_{2}\right)=a^{2} \operatorname{Var}\left(x_{1}\right)+b^{2} \operatorname{Var}\left(x_{2}\right)+2 a b \operatorname{Cov}\left(x_{1}, x_{2}\right)\) c. In Problem \(2.15 \mathrm{d}\) we looked at the variance of \(X=k x_{1}+(1-k) x_{2} \quad 0 \leq k \leq 1 .\) Is the conclusion that this variance is minimized for \(k=0.5\) changed by considering cases where \(\operatorname{Cov}\left(x_{1}, x_{2}\right) \neq 0 ?\) d. The correlation coefficient between two random variables is defined as \\[ \operatorname{Corr}\left(x_{1}, x_{2}\right)=\frac{\operatorname{Cov}\left(x_{1}, x_{2}\right)}{\sqrt{\operatorname{Var}\left(x_{1}\right) \operatorname{Var}\left(x_{2}\right)}} \\] Explain why \(-1 \leq \operatorname{Corr}\left(x_{1}, x_{2}\right) \leq 1\) and provide some intuition for this result. e. Suppose that the random variable \(y\) is related to the random variable \(x\) by the linear equation \(y=\alpha+\beta x\). Show that \\[ \beta=\frac{\operatorname{Cov}(y, x)}{\operatorname{Var}(x)} \\] Here \(\beta\) is sometimes called the (theoretical) regression coefficient of \(y\) on \(x\). With actual data, the sample analog of this expression is the ordinary least squares (OLS) regression coefficient.
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