4.5: Some optimization problems.

In Section 3.5.2 we found that local maxima and minima are often points at which the derivative is zero. The algebraic functions for which we can now compute derivatives have only a finite number of points at which the derivative is zero or does not exist and it is usually a simple matter to search among them for the highest or lowest points of their graphs. Such a process has long been used to find optimum parameter values and a few of the traditional problems that can be solved using the derivative rules of this chapter are included here. More optimization problems appear in Chapter 8 Applications of the Derivative.

Assume for this section only that all local maxima and local minima of a function, \(F\), are found by computing \(F ^{\prime}\) and solving for \(x\) in \(F ^{\prime} (x) = 0\).

Example 4.5.1 A forester needs to get from point \(A\) on a road to point \(B\) in a forest (see diagram in Figure \(\PageIndex{1}\)). She can travel 5 km/hr on the road and 3 km/hr in the forest. At what point, \(P\), should she leave the road and enter the forest in order to minimize the time required to travel from \(A\) to \(B\)?

Figure \(\PageIndex{1}\): Diagram of a forest and adjacent road for Example 4.5.1

Solution. She might go directly from \(A\) to \(B\) through the forest; she might travel from \(A\) to \(C\) and then to \(B\); or she might, as illustrated by the dashed line, travel from \(A\) to a point, \(P\), along the road and then from \(P\) to \(B\).

Assume that the road is straight, the distance from \(B\) to the road is 5 km and the distance from \(A\) to the projection of \(B\) onto the road (point \(Q\)) is 6 km. The point, \(P\), is where the forester leaves the road; let \(x\) be the distance from \(A\) to \(P\). The basic relation between distance, speed, and time is that

\[\text{Distance (km) = Speed (km/hr)} \times \text{Time (hr)}\]

so that

\[\text{Time} = \frac{\text{Distance}}{\text{Speed}}\]

The distance traveled and time required are

\[\begin{array}\
&\text{Along the road} & \text{In the forest}\\
\text{Distance} & x & \sqrt{(6-x)^{2}+5^{2}}\\
\text{Time} & \frac{x}{5} & \frac{\sqrt{(6-x)^{2}+5^{2}}}{3}
\end{array}\]

The total trip time, \(T\), is written as

\[T = \frac{x}{5} + \frac{\sqrt{(6 - x) ^{2} + 5^{2}}}{3} \label{4.12}\]

A graph of \(T ~ vs ~ x\) is shown in Figure \(\PageIndex{2}\). It appears that the lowest point on the curve occurs at about \(x\) = 2.5 km and \(T\) = 2.5 hours.

Figure \(\PageIndex{2}\): Graph of Equation \ref{4.12}; total trip time, \(T\), vs distance traveled along the road, \(x\) before entering the forest.

Explore 4.5.1 It appears that to minimize the time of the trip, the forester should travel about 2.5 km along the road from \(A\) to a point \(P\) and enter the forest to travel to \(B\). Observe that the tangent to the graph at the lowest point is horizontal, and that no other point of the graph has a horizontal tangent.

Compute the derivative of \(T(x)\) for

\[T(x) = \frac{x}{5} + \frac{\sqrt{(6 - x) ^{2} + 5^{2}}}{3}\]

Note: The constant denominators may be factored out, as in

\[[\frac{x}{5}]^{\prime} = [\frac{1}{5} x]^{\prime} = \frac{1}{5} [x] ^{\prime} = \frac{1}{5} \times 1 = \frac{1}{5}\]

You should get

\[T^{\prime}(x)=\frac{1}{5}+\frac{1}{3} \frac{1}{2}\left((6-x)^{2}+5^{2}\right)^{-1 / 2} \times 2(6-x)(-1)\]

Find the value of \(x\) for which \(T ^{\prime} (x) = 0\).

Your conclusion should be that the forester should travel 2.25 km from \(A\) to \(P\) and that the time for the trip, \(T(2.25) = 2.53\bar{3}\) hours.

 

Exercises for Section 4.5, Some optimization problems.

Exercise 4.5.1 In Example 4.5.1, what should be the path of the forester if she can travel 10 km/hr on the road and 4 km/hr in the forest?

Exercise 4.5.2 The air temperature is \(-10 ^{\circ} F\) and Linda has a ten mile bicycle ride from the university to her home. There is no wind blowing, but riding her bicycle increases the effects of the cold, according to the wind chill chart in Figure 4.5.2 provided by the Centers for Disease Control. The formula for computing windchill is

\[\text { Windchill }(\mathrm{F})=35.74+0.6215 T-35.75 V^{0.16}+0.4275 T V^{0.16}\]

where: \(T\) = Air Temperature (F) and \(V\) = Wind Speed (mph).

Assume that if she travels at a speed, \(s\), then she looses body heat a rate proportional to the difference between her body temperature and wind chill temperature for speed \(s\). (Because no wind is blowing, \(V = s\)).

1. At what speed should she travel in order to minimize the amount of body heat that she looses during the 10 mile bicycle ride?
2. Frostbite is skin tissue damage caused by prolonged skin tissue temperature of \(23 ^{\circ} F\). The time for frostbite to occur is also shown in Figure 4.5.2. What is her optimum speed if she wishes to avoid frostbite.
3. Discuss her options if the ambient air temperature is \(-20 ^{\circ} F\).

Figure for Exercise 4.5.2 Table of windchill temperatures for values of ambient air temperatures and wind speeds provided by the Center for Disease Control at http://emergency.cdc.gov/disasters/winter/pdf/cold guide.pdf. It was adapted from a more detailed chart at http://www.nws.noaa.gov/om/windchill.

Exercise 4.5.3 If \(x\) pounds per acre of nitrogen fertilizer are spread on a corn field, the yield is

\[200 - \frac{4000}{x + 25}\]

bushels per acre. Corn is worth $6.50 per bushel and nitrogen costs $0.63 per pound. All other costs of growing and harvesting the crop amount to $760 per acre, and are independent of the amount of nitrogen fertilizer applied. How much nitrogen per acre should be used to maximize the net dollar return per acre? Note: The parameters of this problem are difficult to keep up to date.

Exercise 4.5.4 Optimum cross section of your femur. R. M. Alexander³ has an interesting analysis of the cross section of mammal femurs. Femurs are hollow tubes filled with marrow. They should resist forces that tend to bend them, but not be so massive as to impair movement. An optimum femur will be the lightest bone that is strong enough to resist the maximum bending moment, \(M\), that will be applied to it during the life of the animal.

A hollow tube of mass \(m\) kg/m may be stronger than a solid rod of the same weight, depending on two parameters of the tube, the outside radius, \(R\), and the inside radius, \(x \times R ~ (0 \leq x < 1)\), see Figure \(\PageIndex{2}\). For a given moment, \(M\), the relation between \(R\) and \(x\) is

\[R=\left[\frac{M}{K\left(1-x^{4}\right)}\right]^{\frac{1}{3}}=\left(\frac{M}{K}\right)^{\frac{1}{3}}\left(1-x^{4}\right)^{-\frac{1}{3}} \label{4.13}\]

The constant \(K\) describes the strength of the material.

Figure \(\PageIndex{2}\): Consider a femur to be a tube of radius \(R\) with solid bone between \(kR\) and \(R\) and marrow inside the tube of radius \(kR\).

Let \(\rho\) be bone density and assume marrow density is \(\frac{1}{2} \rho\). Then the mass per unit length of bone, \(m_b\), is

\[\begin{array}\
m_{b} &=\rho \times\left(\pi R^{2}-\pi(R \times x)^{2}\right) \\
&=\rho \pi\left(1-x^{2}\right) R^{2} \\
&=\rho \pi\left(\frac{M}{K}\right)^{\frac{2}{3}}\left(1-x^{2}\right)\left(1-x^{4}\right)^{-\frac{2}{3}}
\end{array} label{4.14}\]

1. Write an equation for the mass per unit length of the bone marrow similar to Equation \ref{4.14}.
2. Let \(m\) be total mass per unit length; the sum of \(m_b\) and the mass per unit length of marrow. We would like to know the derivative of \(m\) with respect to \(x\) for \[m=C\left(\left(1-\frac{1}{2} x^{2}\right)\left(1-x^{4}\right)^{-\frac{2}{3}}\right) . \quad C=\rho \pi\left(\frac{M}{K}\right)^{\frac{2}{3}} \label{4.15}\] You will see in Chapter 6 that \[[m]^{\prime}=C\left(\left[1-\frac{1}{2} x^{2}\right]^{\prime} \times\left(1-x^{4}\right)^{-\frac{2}{3}}+\left(1-\frac{1}{2} x^{2}\right) \times\left[\left(1-x^{4}\right)^{-\frac{2}{3}}\right]^{\prime}\right)\] Finish the computation of \([ m ] ^{\prime}\) and simplify the expression.
3. Find a value of \(\bar{x}\) for which \(x = \bar{x}\) yields \(m^{\prime} = 0\). \[-x\left(1-x^{4}\right)^{-\frac{2}{3}}+\frac{8}{3} x^{3}\left(1-\frac{1}{2} x^{2}\right)\left(1-x^{4}\right)^{-\frac{5}{3}}=0\]
4. The value \(\bar{x}\) computed in Part c. is the x-coordinate of the lowest point of the graph of \(m\) shown in Exercise Figure 4.5.4. Alexander shows the values for \(x\) for five mammalian species; for the humerus they range from 0.42 to 0.66 and for the femur they range from 0.54 to 0.63. Compare \(\bar{x}\) with these values.
5. Exercise Figure 4.5.4B is a cross section of the human leg at mid-thigh. Estimate x for the femur.

Alexander modifies this result, noting that Equation 4.13 is the breaking moment, and a bone with walls this thin would buckle before it broke, and noting that bones are tapered rather than of uniform width.

Figure for Exercise 4.5.4 A. Graph of Equation \ref{4.15}, \(m\), the mass of bone plus marrow, as a function of the ratio, \(x\). Mass per unit length of a solid bone has been arbitrarily set equal to one. B. Cross section of a human leg at mid thigh. http://en.wikipedia.org/wiki/Fascial compartments of thigh. Lithograph plate from Gray’s Anatomy, not copyrightable.

 

³ R. McNeil Alexander, Optima for Animals, Princeton University Press, Princeton, NJ, 1996, Section 2.1, pp 17-22.