7.3 Optimization

7.3.1 Examples of Optimization Problems

We now turn our attention to solve optimization problems. To solve such problems, there are two fundamental steps:

1. Find a function of a single variable together with its domain that models the quantity to be optimized.

   This step, in turn, requires that we focus on two aspects of the problem:

   (QO) quantity to be optimize;    (C) constraints.

   At this point, it would be convenient for the reader to review ?? for examples where this step is performed, and, in particular, to pay attention to the General Observation that outlines how to proceed to find the single variable function modeling the quantity to be optimized.

2. Apply the procedures developed in 7.2 to determine the absolute maximum and absolute minimum values of the function.

Example 1 A rectangle is inscribed in an equilateral triangle of side \(s\) cm as in the picture. Find the area and the dimensions of the largest rectangle that can be thus inscribed in the triangle.

Solution. (a) See Example ?? in ?? for a figure illustrating the problem and where it is show that the quantity to be optimized, which is the area of the inscribed rectangle is given by

\[\hspace{.5in}\hphantom{(*)}
A(b) = b \left(\frac{\sqrt{3}}{2}( s-b)\right)=\frac{\sqrt{3}}{2}b ( s-b), \;\; 0\leq b\leq s.
\hspace{.5in}(*)\]
This is the first part of the optimization process.

(b) Finding the absolute maximum value of the function in \((*)\).

Given that \(A=A(b)\) is a quadratic function, it is continuous and differentiable everywhere. To find the maximum possible area, we can proceed in two different ways: (i) using algebra; (ii) using calculus.
(i) was carried out in Example ?? in ??.

Solution via calculus.

(1) The values at the endpoints are \(A(0)= A(s) = 0\) (the rectangle collapses to a single line).

(2) Values at the critical numbers.
\[A'(b) = \frac{\sqrt{3}}{2} ( s-b) – \frac{\sqrt{3}}{2}b=0 \;\; \Longleftrightarrow \;\; b=\frac{s}{2}. \;\; \Longrightarrow \;\; A\left(\frac{ s}{2}\right)=\frac{\sqrt{3}}{8} s^2.\]

(3) Comparing values, the maximum area is \(\displaystyle A\left(\frac{ s}{2}\right)=\frac{\sqrt{3}}{8} s^2 \text{ cm}^2\).

Example 2An open-top box is to be made from a 20 cm by 32 cm metal-plate by removing a square from each corner of the box and folding up the flaps on each side (see Figure 7.14). What size square should be cut out of each corner to get a box that has the maximum possible volume?

Solution. (a) Find a function of a single variable that models the quantity to be optimizes.

This was accomplished in Example ?? in ??, where it was shown that if \(s\) denotes the length of the sides of the squares being cut out from the corners of the metal-plate, then
\[V= (32-2s) ( 20-2s) s, \;\;\; 0 \leq s \leq 10.\]

(b) Find the maximum volume of the function \(V= (32-2s) ( 20-2s) s, \;\;\; 0 \leq s \leq 10\).

(1) The values at the endpoints are \(V(0)= 0\) (the box collapses to a rectangle).

(2) Values at the critical numbers.
\[V'(s) = 4 (s-4) (3 s-40) =0 \;\; \Longleftrightarrow \;\; s\in \left \{4, \frac{40}{3}\right \}. \;\; \Longrightarrow \;\; V\left(4\right)=12\cdot24\cdot4 = 1152 \left (\text{cm}^3\right ).\]
Note that \(\displaystyle \frac{40}{3} > 10\) is not in the domain of definition of \(V\).

(3) Comparing values, the maximum volume is \(\displaystyle V\left(4\right)= 1152\, \text{cm}^3\), and is attained when the sides of the cut out squares are \(s=4\) cm.

Example 3A cylindrical container with no top is to have a capacity of 1000 \(\pi \; m^3\). What is the least possible amount of material needed to construct such a cylinder?

Solution. (a) Modeling the quantity to be optimized as a function of a single variable.

Quantity to be optimized: minimize the amount of material used to construct the container. Or, equivalently, minimize the surface area of the container.

Constraints: the container is a cylinder with no top and its capacity is \(V= 1000\pi \; m^3\).

Figure 7.13

\[SA= SA_{\text{base}}+SA_{\text{lateral surface}}.\]
Let \(r\) denote the radius of the cylinder, and \(h\) its height. Then, we have

the base of the cylinder has area given by \(SA_{base} = \pi r^2\) m\(^2\);

the lateral surface of the container is given by \(SA_{\text{lateral surface}} = 2\pi r h\) m\(^2\).

To see the validity of this last equation, cut the lateral surface along a vertical height, and flatten the cylinder. The width of the rectangle thus obtained is precisely \(2pi r\) (the perimeter of the base of the cylinder).

It follows that
\[SA= SA_{\text{base}}+SA_{\text{lateral surface}} = \pi r^2 + 2\pi r h.\]
In order to rewrite this last expression for \(SA\), we use the constraint
\[V=1000\pi = \pi r^2h\,(\text{m}^3) \;\;\Longrightarrow \;\; h=\frac{1000}{r^2}.\]
It follows that the surface area can be rewritten in terms of a single variable as follows:
\[SA = \pi r^2+ 2\pi r \frac{1000}{r^2}=\pi r^2+ \frac{2000\pi }{r},\;\; r>0.\]
This completes the modeling of the quantity to be optimized.

(b) Find the minimum possible area.

Note. As the domain of the function is an open interval we apply

(1) Critical numbers.
\[SA'(r)=2\pi r – \frac{2000\pi }{r^2}=2\pi \left(\frac{r^3-1000}{r^2}\right).\]
Thus, the critical numbers are the values $r>0$ such that $SA'(r)=0 $ or $SA'(r)$ undefined.

\(SA'(r)=0 \Longleftrightarrow r^3=1000 \Longleftrightarrow r=10\)
\(SA'(r)\;\; undefined \Longleftrightarrow r=0 \notin D_{SA}\).

Thus, there is only one critical number \(r=10\), and we showed in Example in that \(SA(10)\) is an absolute minimum of \(SA\). Hence
\[SA(10)=\pi (10)^2+ \frac{2000\,\pi}{10}= 300\,\pi \,\text{cm}^2 \;\; \text{(minimum area).}\]

The following example presents two alternative ways to solve the same problem. One uses algebra and the other uses trigonometry. When solving a problem in general, it is always convenient to consider all possible methods available to solve such a problem.

Example 4
A person located at point \(A\) on a riverbank would like to get across to a point \(D\) located 6 km down the river. The river is 2 km wide. If the person can row at a rate of 3 km/h and can walk at a rate of 5 km/h, where should the person land in order to minimize the time to get to \(D\) .

Solve the problem combining calculus with (a) algebra using \(x\) as a variable; (b) trigonometry using \(\theta\) as a variable (see Figure ). (Neglect the speed of the water as the person rows.)

Figure 7.14: Example 4

Solution.
(a) Modeling the quantity to be optimizes by means of a function of a single variable.

This step was completed in Example ?? in ??, where it was show that the time it takes the person to reach point \(D\) is given by

\[t = t(x)
\frac{\sqrt{x^2+2^2}}{3} + \frac{6-x}{5}, \;\; 0\leq x \leq 6.
\]
or
\[t(\theta) = \frac{2\sec \theta}{3}+\frac{6 – 2\tan \theta}{5}, \;\; \theta \in [0, \arctan 3].\]

(b) Find the absolute minimum possible time it can take the person to cross the river.

Solution 1. Algebra.
Here is an outline of the process. Fill in the details. Note that \(t\) is continuous on \([0,6]\).

\[t'(x)= \frac{x}{3\,\sqrt{4+x^2}}-\frac{1}{5}. \]
\(t'(x)\) is well defined for all \(x \in [0,6]\), thus, the critical numbers arethe numbers in \((0,6)\) such that \(t'(x)=0\).
\begin{align*}
t'(x)=0 & \; \Longleftrightarrow \; \frac{x}{3\,\sqrt{4+x^2}}-\frac{1}{5} = 0
\; \Longleftrightarrow \; \frac{5x -3\,\sqrt{4+x^2} }{15\,\sqrt{4+x^2}} = 0 \; \Longleftrightarrow \; 5x -3\,\sqrt{4+x^2} = 0 \; \Longrightarrow\\
& \; \Longrightarrow \; 25x^2 = 9(4+x^2) \; \Longleftrightarrow \; 16x^2= 36 \; \underset{?}{\Longrightarrow} \; x=\frac{3}{2}.
\end{align*}
The times at the critical point and the endpoints are
\[ t\left (\frac{3}{2}\right ) = \frac{26}{15}; \;\; t(0)= \frac{28}{15}; \;\; t(6) = \frac{2\sqrt{10}}{3}.\]
Since \(\sqrt{10} > 3\), it follows that the person will take the minimum possible time to reach point \(D\) if \(C\) is located \(\displaystyle \frac{3}{2}\) km away from \(B\).

Solution 2. Trigonometry.
\[t'(\theta) = \frac{2\sec \theta\tan \theta}{3}-\frac{ 2\sec^2 \theta}{5}
=\frac{2\sec \theta}{15}(5\tan \theta -3\sec \theta).\]
Note that \(t’\) is well defined on the domain of \(t\). Thus, the critical numbers are the values of \(\theta\) in \((0, \arctan 3)\) such that \(t'(\theta) = 0\).
\[t'(\theta)=0 \; \underset{?}{\Longleftrightarrow} \; 5\tan \theta -3\sec \theta =0
\; \underset{?}{\Longleftrightarrow} \; 5\sin \theta -3 =0 \; \Longleftrightarrow \; \sin \theta =\frac{3}{5} \; \Longrightarrow \; \theta = \arcsin \frac{3}{5}.\]
Note that
\[\alpha = \arcsin \frac{3}{5} \; \Longleftrightarrow
\left \{\renewcommand{\arraystretch}{2}\begin{array}{l}
\displaystyle \sin \alpha = \frac{3}{5}\\ \displaystyle -\frac{\pi}{2}\leq \alpha\leq \frac{\pi}{2}
\end{array}\right .
\;\Longrightarrow \; \cos^2\alpha = \frac{16}{25} \;\Longrightarrow \; \sec^2\alpha = \frac{25}{16} \;\Longrightarrow\]
\[
\;\Longrightarrow \; \tan^2\alpha =\sec^2\alpha – 1 = \frac{25}{16}- 1 = \frac{9}{16}
\;\Longrightarrow \; \tan \alpha = \frac{3}{4} < 3 \; \Longrightarrow \; \arcsin \frac{3}{5} \in (0, \arctan 3). \] We need to compare the values of \(t\) at the critical point and at the endpoints. In the notation of the above calculation, we have \[t(\alpha) = \frac{2\sec \alpha}{3}+\frac{6 - 2\tan \alpha}{5} = \frac{\displaystyle 2\cdot\frac{5}{4}}{3}+\frac{\displaystyle 6 - 2\cdot\frac{3}{4}}{5} = \frac{26}{15}.\] We leave it as an exercise to show that \(t(\arctan 3)\) is as stated below. \[ t\left (\arcsin \frac{3}{5}\right )=\frac{26}{15}, \;\; t(0)= \frac{2}{3}+\frac{6}{5}= \frac{28}{15}, \;\; t(\arctan 3) \underset{?}{=} \frac{2\sqrt{10}}{3}.\] These values are the same as above and those obtained using algebra, and the person will take the minimum time to reach point \(D\) when the angle \(\displaystyle \theta = \arcsin (3/5)\) radians.

Example 5 Let \(\triangle ABC\) be a right triangle with \(\angle C = 90^{\circ}\). Let \(\triangle ADE\) be a second right triangle with \(\angle E =90^{\circ}\), \(|EC|=4\) and \(|DE|=4\) (see Figure 7.15(a)). Find the dimensions of \(\triangle ABC\) such that \(|AB|\) has minimum length. Solve this problem in two ways, combining calculus with (a) algebra; (b) trigonometry.

Figure 7.15: Example 5: (a)

Figure 7.15: Example 5: (b)

Solution.
(a) Find a function of a single variable that models the quantity to be optimizes.

In this case, the target function is the the length \(|AB|\). The constraints are the similar right triangles and the fixed lengths \(|EC|=4\) and \(|DE|=4\).

Using similar triangles, we have
\[\frac{|AB|}{|AC|}= \frac{|AD|}{|AE|} \;\; \Longleftrightarrow \;\; |AB| = \frac{|AD|}{|AE|} |AC| \;\; \text{ and } \;\; |AC| = |AE|+4.\]

We now introduce the following notation: \(x=|AE|\) and \(\theta = \angle DAE\) (see Figure 7.15)(b).

Solution 1. Algebra. Using the Pythagorean theorem
\[|AB|= \frac{|AD|}{|AE|} |AC| =\frac{\sqrt{x^2+4^2}}{x}(x+4).\]
Note that necessarily, \(x > 0\). Thus, the target function is given by
\[f(x) =\frac{x+4}{x}\sqrt{x^2+4^2} , \;\; x\in (0, \infty).\]

Solution 2. Trigonometry.
\[|AB|= \frac{|AD|}{|AE|} |AC|
=\frac{4\csc\theta}{4\cot \theta}\left (4\cot \theta + 4\right ) = \sec\theta\left (4\cot \theta + 4\right ).\]
Note that in this case, \(\displaystyle 0 < \theta < \frac{\pi }{2}\). Thus, the target function in terms of \(\theta\) is given by \[f(\theta) = 4\sec\theta\left (\cot \theta + 1\right ), \;\; \theta \in \left (0 ,\frac{\pi}{2}\right ).\] (b) Find the absolute minimum value of the target function. Solution 1. Algebra. Verify that for \(f\) as above
\[f'(x) = \frac{x^3-64}{x^2\sqrt {x^2 + 16}} =
\frac{\left (x -4\right )\left (x^2 +4 x+ 16\right )}{x^2\sqrt {x^2 + 16}}, \;\; x > 0\]
(explain why). And
\[f'(x) = 0 \;\; \Longleftrightarrow \;\; x = 4.\]
Thus, \(f\) has a single critical number. Furthermore
\[f'(x) < 0, \; \text{ if }\; 0 < x < 4, \;\; \text{ and } \;\; f'(x) > 0, \text{ if }\;4 < x < \infty.\] It follows that, by Proposition 3(a),
\[f\left (4\right ) =\frac{4+4}{4}\sqrt{4^2+4^2} = 8\sqrt{2}\; (\text{m})\]
is the absolute minimum value of \(\displaystyle f(x) =\frac{x+4}{x}\sqrt{x^2+4^2}\) , \( x\in (0, \infty)\). That is, the minimal length of \(|AB|\).

Solution 2. Trigonometry. In this case,
\(f(\theta) = 4\sec\theta\left (\cot \theta + 1\right )\), \(\displaystyle \theta \in \left (0 ,\frac{\pi}{2}\right )\). Thus
\begin{align*}
f'(\theta) &
=4 \sec\theta\tan \theta\left (\cot \theta+1\right ) + 4\sec\theta\left (-\csc^2 \theta \right )
=4 \sec\theta\left (\tan \theta\left (\cot \theta+1\right ) -\csc^2 \theta \right) = \\
&=
4 \sec\theta\left ( 1+ \frac{1}{\cot \theta} -\csc^2 \theta \right) \underset{?}{=}
4 \sec\theta\left (\frac{1}{\cot \theta} -\cot^2 \theta \right) =
4 \sec\theta\tan \theta\left (1 -\cot^3 \theta \right).
\end{align*}
It follows that for \(\displaystyle \theta \in \left (0 ,\frac{\pi}{2}\right )\)
\[f'(\theta)= 0\;\; \Longleftrightarrow \;\; \cot^3 \theta = 1 \;\; \Longleftrightarrow \;\; \cot \theta = 1 \;\; \Longleftrightarrow \;\; \theta = \frac{\pi}{4}.\]
Thus, \(f\) has a single critical number. Furthermore, given that \(y = \cot \theta\) is decreasing,we have
\[f'(\theta) < 0, \; \text{ if }\; 0 < \theta < \frac{\pi}{4}, \;\; \text{ and } \;\; f'(\theta) > 0, \; \text{ if }\; \frac{\pi}{4} < \theta < \frac{\pi}{2}\] (explain why). It follows that, by Proposition 3(a),
\[f\left (\frac{\pi}{4}\right ) = 4\sec\frac{\pi}{4}\left (\cot \frac{\pi}{4}+ 1\right ) = 4\,\sqrt{2} \,(1+1) = 8 \,\sqrt{2} \; (\text{m})\]
is the absolute minimum value of \(f(\theta) = 4\sec\theta\left (\cot \theta + 1\right )\), \(\displaystyle \theta \in \left (0 ,\frac{\pi}{2}\right )\) (as expected).

7.3.2 Exercises

1. A rectangle is inscribed in an isosceles triangle with its base along the base of the triangle. If the the base of the triangle measures 20 cm and the height 30 cm. Find the area of the largest rectangle that can be thus inscribed in the triangle.
2. A piece of wire of length 4 m is to be cut into two pieces. One piece is bent into a square, and the other into a circle. Where should the wire be cut (or not cut) to enclose (a) the maximum; (b) the minimum possible area by the shape(s). Determine the exact area in each case.
3. A rectangular field that borders a straight river on one side is to be delimited by a fence with no fencing needed along the river. The field is to be divided into three parts with fencing perpendicular to the river. The fencing to delimit the filed is of heavy duty and costs $12 per meter, while the fencing to subdivide the field only costs $4 per meter. Find the dimensions of the maximum possible area for the field if the budget for the fencing is $7,200.
4. A box with no lid and a square base is to have a volume of 8 m\(^3\). What is the least possible amount of material needed to construct such a box?
5. To construct a box with lid and a square base there are available 16 m\(^2\) of cardboard. What is the maximum possible volume of such a box?
6. A rectangle is inscribed in a circle of radius \(R\). Find the area and the dimensions of such a rectangle which has the largest area.

Solve this problem in two different ways (see Figure 7.16 # 6), using calculus and (a) algebra with \(w\) as a variable; (b) trigonometry with \(\theta\) as a variable.

7. A right circular cone is to be made from a circular piece of paper of radius 10 cm by cutting a circular sector (see Figure 7.16 # 7), and gluing the edges of the remaining piece of paper. Find the maximum possible volume of such a cone.


Figure 7.16: # 6



Figure 7.16: # 7



Figure 7.16: # 9

8. An open-top box is to be made from a 30 cm by 80 cm metal-plate by removing a square from each corner of the box and folding up the flaps on each side. What size square should be cut out of each corner to get a box that has the maximum possible volume?
9. An isosceles trapezoid is inscribed in a semicircle of radius 2 m so that its largest base is along the diameter. Find the maximum possible area of the trapezoid.

Solve this problem in two different ways (see Figure 7.16 # 9), using calculus and

(a) algebra with \(x\) as a variable; (b) trigonometry with \(\theta\) as a variable.

10. An internet company is located at the north bank of a river that is 3 km wide. They plan to set up a server located 10 km east on the south bank of the river at a point C. For this, the will install fiber-optic cable to connect the new server with their main server. First they will connect the main server with a substation at a point B on the north bank, and from B, they will connect with the new server at C. It is three times more expensive to lay out the cable underwater than on land. The cost on land is $4,000 per km. Where should the company locate the substation B to minimize the cost of laying out the cable?

Solve this problem in two different ways (see Figure 7.17 # 10), using calculus and

(a) algebra with \(x\) as a variable; (b) trigonometry with \(\theta\) as a variable.



Figure 7.17: # 10



Figure 7.17: # 12



Figure 7.17: # 12

11. Two poles are connected by a taut wire that is also connected to the ground. One pole is 3 m tall and the other is 6 m tall. There is a distance of 10 m between the two poles.

    (a) Where should the wire be anchored to the ground to minimize the amount of wire needed? (b) Determine the amount of wire needed in this case.

Solve this problem in two different ways (see Figure 7.17 # 11), using calculus and

(a) algebra with \(x\) as a variable; (b) trigonometry with \(\theta\) as a variable.

12. Find the exact value of the area of the isosceles trapezoid with congruent sides 8 cm long, and a base also 8 cm long (see Figure 7.17 # 12) of maximum area.

Solve this problem in two different ways, using calculus and (a) algebra with \(x\) as a variable; (b) trigonometry with \(\theta\) as a variable.

(a) algebra with \(x\) as a variable; (b) trigonometry with \(\theta\) as a variable.

13. A construction company needs to perform a study of a high-rise building located 8 ft behind an 27 ft tall wall (see Figure 7.18 # 13). What is the necessary minimum length of a plank to reach the building over the wall.

Solve this problem in two different ways, using calculus and (a) algebra; (b) trigonometry.

14. A girl located at point \(A\) (see Figure 7.18 # 14) of a circular swimming pool with diameter 20 m wants to go to point \(C\) located diametrically opposite to \(A\). The girl can walk twice as fast as she can swim. If she can swim at a rate of \(\frac{1}{2}\) m/sec, determine the trajectory that will take her to reach \(C\) in (a) the least amount of time; (b) the longest amount of time.
    

    Figure 7.18: # 13

    

    Figure 7.18: # 14

    

    Figure 7.18: # 15

15. Find the dimensions and the maximum area of an isosceles triangle inscribed in a circle of radius \(R\) m (see Figure 7.18 # 15). Solve this problem in two different ways, using calculus and (a) algebra; (b) trigonometry.
16. A cylinder is inscribed in a sphere of radius \(R\).
1. Find the maximum possible surface area of such a cylinder.
2. Find the maximum possible volume of such a cylinder.

Solve this problems in two different ways, using calculus and (a) algebra; (b) trigonometry.