In a great many practical problems, we want to find the least amount of time, the least cost, the greatest benefit, optimum size, etc. In such problems, which are called **optimization problems**, we look for the maximum or minimum value of a function and the particular value of the variable that gives such a value. In this section, we outline a strategy to systematically attack optimization problems by using differential calculus.

The graph of \(f\) is shown in Figure 1.

Because \(0\leq x\leq4\), the problem reduces to maximization of \(A(x)\) on the interval \([0,4]\). \[A'(x)=\sqrt{16-x^{2}}+x\frac{-2x}{2\sqrt{16-x^{2}}}.\] Setting \(A'(x)=0\), we obtain \[\sqrt{16-x^{2}}-\frac{x^{2}}{\sqrt{16-x^{2}}}=0\] multiplying both sides by \(\sqrt{16-x^{2}}\) \[16-x^{2}-x^{2}=0\] \[\Rightarrow x=\pm\sqrt{8}=\pm2\sqrt{2}.\] Only \(x=2\sqrt{2}\) is in the interval \([0,4]\). Therefore, the absolute maximum of \(A\) is one of the following numbers \[A(0)=0,\quad A(4)=0\] and \[A(2\sqrt{2})=\sqrt{8}\sqrt{16-8}=8.\] The above calculations show us that the rectangle of maximum area inscribed in the circle is a square of length \(\sqrt{8}=2\sqrt{2}\) and of area 8.

The graph of \(A\) is shown in Figure 3.

\(x=\) length of each side of the square

\(V=\) volume of the open box

We need to find a relationship between \(V\) and \(x\). Because the dimensions of the box are \(48-2x\), \(18-2x\), and \(x\) (Figure 6), the volume of the box is \[V=(48-2x)(18-2x)x=4x^{3}-132x^{2}+864x.\]

Because the dimensions of the box must be nonnegative, we have \[\begin{cases} 0\leq x\\ 0\leq18-2x & \Rightarrow\\ 0\leq48-2x \end{cases}0\leq x\leq9.\] Because \(V\) is continuous on the interval \([0,9]\), its absolute maximum occurs either at an endpoint \(x=0\), \(x=9\), or at a critical point. To obtain critical points \[\frac{dV}{dx}=12x^{2}-264x+864=0\] \[\Rightarrow x^{2}-22x+72=(x-4)(x-18)=0\] \[\Rightarrow x=4,\text{ or }x=18.\] However, \(x=18\not\in[0,9]\). Therefore, the only critical point in the interval \([0,9]\) is \(x=4\), and the absolute maximum of \(V\) is one of the following numbers.

\(x\) | \(0\) | \(4\) | \(9\) |
---|---|---|---|

\(V(x)\) | \(0\) | \(1600\) | \(0\) |

Therefore, the maximum volume is obtained when \(x=4\).

The graph of \(V(x)\) is shown in Figure 6.

\(x\) | \(-\infty\) | \(-\sqrt{2}\) | \(0\) | \(\sqrt{2}\) | \(+\infty\) | ||||
---|---|---|---|---|---|---|---|---|---|

sign of \(x+\sqrt{2}\) | \(—\) | \(0\) | \(+++\) | \(+\) | \(+++\) | \(+\) | \(+++\) | ||

sign of \(x\) | \(—\) | \(-\) | \(—\) | \(0\) | \(+++\) | \(+\) | \(+++\) | ||

sign of \(x-\sqrt{2}\) | \(—\) | \(—\) | \(—\) | \(+\) | \(+++\) | ||||

\(\therefore\)sign of \(f^\prime(x)\) | \(—\) | \(0\) | \(+++\) | \(0\) | \(—\) | \(0\) | \(+++\) | ||

Increasing/Decreasing \(f(x)\) | \(\searrow\) | \(\nearrow\) | \(\searrow\) | \(\nearrow\) |

reveals that \(x=0\) yields a local maximum, whereas \(x=\pm\sqrt{2}\) yield a minimum distance. \[\left.d\right|_{x=\pm\sqrt{2}}=\sqrt{2+(2-\frac{1}{2}2)^{2}}=\sqrt{3}.\]

Let’s see what will happen if we use the Second Derivative Test instead of the First Derivative Test \[f^{\prime\prime}(x)=x^{2}-2\] Because \(f^{\prime\prime}(0)=-2<0\), \(f\) has a local maximum at \(x=0\). Because\(f^{\prime\prime}(\pm\sqrt{2})=0\), the Second Derivative Test fails, and we cannot determine if \(f\) has a local maximum, a local minimum or neither at \(x=\pm\sqrt{2}\).

Therefore, \(f(\pm\sqrt{2})=\sqrt{3}\) is the shortest distance from the point \((0,2)\) to the parabola \(8-x^{2}=2y\), and the closest points on the parabola to \((0,2)\) are \((\sqrt{2},3)\) and \((-\sqrt{2},3)\).

The graph of \(d=\sqrt{x^{2}+(2-x^{2}/2)^{2}}\) is shown in Figure 8.

Let’s summarize the strategy that we have applied to solve the examples of this section:

**Strategy for Solving Optimization Problems**

- Identify the dependent variable that needs to be maximized or minimized and every variable that plays a role in the problem. Assign appropriate letters to the variables and constants that remind you their actual meaning.
- Write a primary equation that relates the variable which is to be optimized to the other variables.
- If the primary equation is a function of more than one independent variable, eliminate the extra variables and reduce the equation to one having a single independent variable. This may involve the use of secondary equations relating the independent variables of the primary equation. The discovery of the secondary equations is often facilitated by drawing a figure or two.
- Determine the domain of the independent variable. This domain may be smaller than the natural domain of the function because of limitations inherent to the problem.
- Test the critical points and endpoints of the domain. Use the First Derivative Test or the Second Derivative Test to classify the critical points.

For example, in the last example, the primary equation is \(d=\sqrt{x^{2}+(y-2)^{2}}\). We need to minimize \(d\) (or \(d^{2}\)) but \(d\) is a function of \(x\) and \(y\). So we had to eliminate one of them using a secondary equation. Because \((x,y)\) is on the parabola, the secondary equation is the equation of parabola relating \(x\) and \(y\), which is \(2y=8-x^{2}\).

\(r=\) radius of the cylinder

\(h=\) height of the cylinder

\(V=\) volume of the cylinder

We know \[V=\pi r^{2}h,\tag{i}\] and we want to maximize \(V\) provided that the cylinder is inscribed in the cone. Here \(V\) depends both on \(r\) and \(h\) and it is necessary to express \(V\) in only one variable (either \(r\) or \(h\)). To do so, we look for a relationship that connects \(r\) and \(h\). If we look at the cone from the side, we realize that \(\triangle OAB\) and \(\triangle O’AB’\) are similar (Figure 10).

\[\overset{\triangle}{OAB}\sim\overset{\triangle}{O’AB’}\] \[\Rightarrow\frac{AO’}{AO}=\frac{O’B’}{OB}\] or \[\frac{30-h}{30}=\frac{r}{18}\] \[\Rightarrow30(18)-18h=30r\] \[h=30-\frac{5}{3}r\tag{ii}\] Substituting (ii) into (i), we obtain \[V=\pi r^{2}\left(30-\frac{5}{3}r\right)=30\pi r^{2}-\frac{5}{3}\pi r^{3}.\] Now \(V\) is a function of \(r\) alone. Because \(r\) may vary only between 0 and 18 (the radius of the cone) \[0\leq r\leq18,\] and \(V\) is a continuous function, the existence of a maximum value is guaranteed. The maximum value occurs either at \(r=0\), \(r=18,\) or at a critical number. \[\begin{aligned} V'(r) & =60\pi r-5\pi r^{2}\\ & =5\pi r(12-r).\end{aligned}\] The solutions of \(V'(r)=0\) are \[r=0,\text{ }r=12.\] It suffices to compute \(V(r)\) at the critical numbers and the endpoints

\(r\) | \(0\) | \(12\) | \(18\) |
---|---|---|---|

\(V\) | \(0\) | \(1440\pi\) | \(0\) |

So the maximum value of volume occurs when \(r=12\) and \(h=30-\frac{5}{3}(12)=10\).

The graph of \(V(r)\) is shown in Figure 11.

(a) Find the width of the part folded over when the length of the crease is a minimum,

(b) Find the width when the area folded over is a minimum.

\(x=\) length of \(BC\)

\(y=\) length of \(CE\)

\(L=\) length of \(BE\)

From \(\triangle BCE\) \[L^{2}=x^{2}+y^{2}.\tag{i }\] We want to minimize \(L\), but \(L\) depends on two variables \(x\) and \(y\). To eliminate one of the variables, we need a relationship between \(x\) and \(y\). From \(\triangle BAD\), we have \[AB^{2}+AD^{2}=BD^{2}\] because before folding the page, \(\triangle BED\) was \(\triangle BEC\), we have \(BD=BC\) and thus \[(a-x)^{2}+AD^{2}=x^{2}\] \[\Rightarrow AD^{2}=x^{2}-(a-x)^{2}=2ax-a^{2}\] Because \(AD=CD’=2ax-x^{2}\), from \(\triangle DD’E\) \[DD’^{2}+D’E^{2}=DE^{2}\] and \[a^{2}+(y-AD)^{2}=y^{2}\] or \[a^{2}+(y-\sqrt{2ax-a^{2}})^{2}=y^{2}\] \[\Rightarrow-2y\sqrt{2ax-a^{2}}+2ax=0\] \[\Rightarrow y^{2}=\frac{a^{2}x^{2}}{2ax-a^{2}}=\frac{ax^{2}}{2x-a}\tag{ii}\] Substituting (ii) in (i) we obtain \[L^{2}=x^{2}+\frac{ax^{2}}{2x-a}.\] Instead of minimizing \(L\), we can minimize \(L^{2}\) to avoid differentiation of a radical. Let \[\begin{aligned} f(x) & =x^{2}+\frac{ax^{2}}{2x-a}\\ & =\frac{2x^{3}-ax^{2}+ax^{2}}{2x-a}\\ & =\frac{2x^{3}}{2x-a}.\end{aligned}\] The critical number (or critical point) of \(f\) is obtained by setting \(f^\prime(x)=0\): \[f^\prime(x)=\frac{6x^{2}(2x-a)-4x^{3}}{(2x-a)^{2}}=0\] \[\Rightarrow8x^{3}-6ax^{2}=2x^{2}(4x-3a)=0\]

\[\Rightarrow x=0,x=\frac{3a}{4}.\] Note that \(x\) must be larger than \(a/2\), otherwise the expression under the radical \(L=\sqrt{2x^{3}/(2x-a)}\) becomes negative. Therefore, \(x=0\) is not acceptable. Because \((2x-a)^{2}>0\), the sign of \(f^\prime(x)\) is the same as the sign of its numerator, and the numerator \(8x^{3}-6ax^{2}=2x^{2}(4x-3a)\) is negative if \(x<3a/4\) and is positive if \(x>3a/4\). Therefore, by the First Derivative Test, \(x=3a/4\) is a local minimum point (\(f\) varies from a decreasing function to an increasing one), and \(L\) for \(x=3a/4\) is \[\left.\sqrt{\frac{2x^{3}}{2x-a}}\right|_{x=\frac{3a}{4}}=\frac{3\sqrt{3}}{4}a\approx1.299a.\] If \(x=a\), then \[L=\sqrt{\frac{2a^{2}}{2a-a}}=\sqrt{2}a\approx1.414a\] and \[\lim_{x\to a/2^{+}}L=\lim_{x\to a/2^{+}}\sqrt{\frac{2x^{3}}{2x-a}}\stackrel{\left[\frac{2a^{3}}{0^{+}}\right]}{=}+\infty.\] These calculations show that the absolute minimum of \(L\) is \(3\sqrt{3}a/4\).

The graph of \(L/a\) versus \(x/a\) is shown in Figure 14. \[\begin{aligned} \frac{L}{a} & =\frac{1}{a}\sqrt{\frac{2x^{3}}{2x-a}}\\ & =\sqrt{\frac{2x^{3}}{a^{3}(2\frac{x}{a}-1)}}\\ & =\sqrt{\frac{2(x/a)^{3}}{2(x/a)-1}}\end{aligned}\]

**(b)** The folded area \(A\) is \[A=xy\tag{iii }\] Substituting (ii) in (iii), we obtain \[\begin{aligned} A & =x\sqrt{\frac{ax^{2}}{2x-a}}\\ & =\sqrt{\frac{ax^{4}}{2x-a}}.\end{aligned}\] Again to avoid differentiation of a radical, we can minimize \(A^{2}\) instead of minimizing \(A\). So we minimize \[g(x)=\frac{ax^{4}}{2x-a}.\] The critical point is obtained by setting \[g^\prime(x)=\frac{4ax^{3}(2x-a)-2ax^{4}}{2x-a}=0\]

\[\Rightarrow2ax^{3}\left[2(2x-a)-x\right]=0\] \[\Rightarrow x=0,x=\frac{2a}{3}.\] Again \(x=0\) is not within the domain of \(A\) which is the interval \((a/2,a]\). When \(x=2a/3\), the area is \[A=\frac{2a}{3}\sqrt{\frac{a\frac{4}{9}a^{2}}{\frac{4a}{3}-a}}=\frac{4a^{2}}{3\sqrt{3}}.\] We can show that \(A\to\infty\) as \(x\to+\infty\) or \(x\to a/2^{+}\). Also \[\left.A\right|_{x=a}=a^{2}.\] Therefore, \(A\) has an absolute minimum when \(x=2a/3\).

The graph of \(A/a^{2}\) versus \(x/a\) is shown in Figure 15.

From \(\triangle ABC\): \[\tan\theta=\frac{b}{x}\] and from \(\triangle ADE\): \[\tan\theta=\frac{y}{a}.\] So the relationship between \(x\) and \(y\) is \[\frac{b}{x}=\frac{y}{a}\Rightarrow y=\frac{ab}{x}.\] Therefore, the length of the ladder can be expressed as \[\begin{aligned} L & =\sqrt{x^{2}+b^{2}}+\sqrt{\frac{a^{2}b^{2}}{x^{2}}+a^{2}}\\ & =\sqrt{x^{2}+b^{2}}+\sqrt{\frac{a^{2}b^{2}+a^{2}x^{2}}{x^{2}}}\\ & =\sqrt{x^{2}+b^{2}}+\frac{a}{x}\sqrt{b^{2}+x^{2}}\\ & =\sqrt{x^{2}+b^{2}}\left(1+\frac{a}{x}\right).\end{aligned}\] The critical point of \(L\) is obtained by setting \(dL/dx=0\): \[\begin{aligned} \frac{dL}{dx} & =\frac{2x}{2\sqrt{x^{2}+b^{2}}}\left(1+\frac{a}{x}\right)-\frac{a}{x^{2}}\sqrt{x^{2}+b^{2}}\\ & =\frac{1}{\sqrt{x^{2}+b^{2}}}\left[x+a-\frac{a}{x^{2}}(x^{2}+b^{2})\right]\end{aligned}\] \[\frac{dL}{dx}=0\Rightarrow x-\frac{ab^{2}}{x^{2}}=0\Rightarrow x=\sqrt[3]{ab^{2}}.\] and the minimum length \(L\) is \[\begin{aligned} \left.\sqrt{x^{2}+b^{2}}\left(1+\frac{a}{x}\right)\right|_{x=\sqrt[3]{ab^{2}}} & =\sqrt{a^{2/3}b^{4/3}+b^{2}}\left(1+\frac{a^{2/3}}{b^{2/3}}\right)\\ & =\sqrt{\frac{a^{2/3}b^{4/3}+b^{2}}{b^{4/3}}}\left(b^{2/3}+a^{2/3}\right)\\ & =(a^{2/3}+b^{2/3})^{3/2}.\end{aligned}\] Obviously as \(x\to0\) or \(x\to+\infty\), \(L\to+\infty\), and the local minimum of \(L\) is also its absolute minimum.