Chapter 1: Answers 3 Jack K. Cohen Colorado School of Mines

  1. The Alpha lettuce patch:
    1. `Best' means maximum area (usually).
    2. Intuitively a square would give the maximum area among all rectangular patches. Thus the side (x) is 100/4 = 25 m and the maximum area is 625 m2.
    3. Your figure should have shown a rectangle with two facing sides labeled with x and the other two with 100 - 2x = 2(50 - x) .
    4. x is one side of the rectangle, A is its area. The valid x-values (the ``domain'' of A) are [0, 50] because outside this interval a side will assume a non-physical negative value. We know that even the endpoint values 0 and 50 can't be right since then the area is zero, but it does no harm to include these and we'll see later that posing maximum or minimum problems on a closed interval has some theoretical advantages.
    5. The values in the middle of the most refined Table are:
              
              24.7   624.91
              24.8   624.96
              24.9   624.99
              25.    625.
              25.1   624.99
              25.2   624.96
              25.3   624.91
      

    6. See Figure 1.
      Figure: Graph of A(x) in the critical region of Problem 1.
      \begin{figure}
\epsfysize 100pt
\centerline{\epsffile{ans3p1.eps}}
\end{figure}

    7. The maximizing x is half-way between the zeroes, that is x = 25. One way to justify this is to multiply out the polynomial and complete the square: A(x) = 50x - x2 = - (x - 25)2 + 625. This form of the equation shows clearly that A≤625 for all x and actually achieves the maximum value A = 625 at x = 25. Note: the ``between the zeroes idea'' is limited to quadratics—another and far more general answer to this question is that the maximum occurs where the tangent to the function is horizontal. The pursuit and refinement of this insight is a large part of this semester's work. At the present time, this criteria is not practical for us, since we don't yet know how to find the x's that make the tangent horizontal—but we soon will! (No credit for using this method now, if you happen to have covered this notion in high school.)

    8. In ``real life'' any number of other restrictions might apply. For example, the need for our irrigation lines to be able to reach all parts of the patch.

    9. Still labeling one side as x, we find that the adjacent side has length P/2 - x, so that the area is A(x) = x(P/2 - x). Using the ``half-way between the zeroes'' insight, we intuit that the maximizing x = P/4 and that the corresponding area is A = P2/16. Completing the square verifies this intuition: A(x) = - (x - P/4)2 + P2/16. In most problems, we would not be able to handle the parameter so cleanly. In the absence of some theoretical result or blinding flash of insight, we might well pick some ``typical'' P-values and use graphical/tabular or other methods to attempt to form a hypothesis about the general case.

  2. From Figure 2 (you did draw a figure, didn't you?), we see that now A(x) = 2x(50 - x). Using the ``half-way between'' insight for parabolas or some other method, find that the maximizing x = 25 and the maximum value is A = 1250 m2 (exactly twice as large as for the 4-sided case). For the general case, the formula is A(x) = 2x(P/2 - x) with maximum at x = P/4 of size A = P2/8.

    Figure: Diagram for Problem 2.
    \begin{figure}
\epsfysize 100pt
\centerline{\epsffile{ans3p2.eps}}
\end{figure}

  3. Note that now, we want to minimize the perimeter. From Figure 3, we see that the sum of side lengths is L = x + 1000/x + x = 2x + 1000/x.

    Figure: Diagram for Problem 3.
    \begin{figure}
\epsfysize 100pt
\centerline{\epsffile{ans3p3a.eps}}
\end{figure}
    One mode of solution is a graph. Figure 4 shows a zoom of the critical region obtained (after some experimentation) by using the command

    Plot[2x + 1000/x, {x, 10, 50}, PlotRange -> {80, 150}]

    Picking the point at the minimum gives x $\approx$ 22. Then the command

    Table[{x, 2x + 1000/x}, {x, 22.2, 22.4, 0.05}] //TableForm

    lets us home in on the answer:

    22.2    89.445
    22.25   89.4438
    22.3    89.443
    22.35   89.4427
    22.4    89.4429
    
    Thus x $\approx$ 22.35 and A $\approx$ 89.4427 at the minimum. Calculus methods let use determine that the exact answer is x = 2⋅53/2 $\approx$ 22.3606797749979, giving a minimum of A $\approx$ 89.4427190999916.

    Figure: Plot of 2x + 1000/x in the critical region.
    \begin{figure}
\epsfysize 90pt
\centerline{\epsffile{ans3p3b.eps}}
\end{figure}

  4. A diagram is shown in Figure 5. The unknown is the radius r. As shown, the quantities h = 10 and s = 1000 are given.
    1. See Figure 5.

      Figure: Diagram for Problem 4.
      \begin{figure}
\epsfysize 90pt
\centerline{\epsffile{ans3p4.eps}}
\end{figure}
    2. Use the right triangle to derive: cos(s/r) = r/(r + h), here h = 20, s = 1000.