8.3 Max/min examples

Factor the derivative.

The global maximum is $f(-1) = 6$, the global minimum is $f(-2) = -20$.

We compute $f'(x)=5\,x^4 - 5$, which means that $f(x)$ has no singular points (i.e., it is differentiable for all values of $x$), but it has two critical points:

Note, however, that $1$ is not in the interval $[-2,0]$.

The global maximum and the global minimum for $f(x)$ on the interval $[-2,0]$ will occur at $x=-2$, $x=0$, or $x=-1$.

So, the global maximum is $f(-1) = 6$ while the global minimum is $f(-2) = -20$.

Remember to test endpoints.

Global maximum is $f(2) = 12$, global minimum is $f(1) = -14$.

We compute $f'(x)=5x^4-5$, which means that $f(x)$ has no singular points (i.e., it is differentiable for all values of $x$), but it has two critical points:

Note, however, that $-1$ is not in the interval $[0,2]$.

The global maximum and the global minimum for $f(x)$ on the interval $[0,2]$ will occur at $x=2$, $x=0$, or $x=1$.

So, the global maximum is $f(2) = 12$ while the global minimum is $f(1) = -14$.

Global maximum is $f(4) = 30$, global minimum is $f(2) = -10$.

We compute $f'(x)=6x^2 - 12x = 6x(x-2)$, which means that $f(x)$ has no singular points (i.e., it is differentiable for all values of $x$), but it has the two critical points: $x=0$ and $x=2$. Note, however, $0$ is not in the interval $[1,4]$.

So, the global maximum is $f(4) = 30$ while the global minimum is $f(2) = -10$.

One way to decide whether a critical point $x=c$ is a local extremum is to consider the first derivative. For example: if $f'(x)$ is negative for all $x$ just to the left of $c$, and positive for all $x$ just to the right of $c$, then $f(x)$ decreases up till $c$, then increases after $c$, so $f(x)$ has a local minimum at $c$.

Local max at $(-2,20)$, local min at $(2,-12)$.

Since $h(x)$ is a polynomial, it has no singular points. We compute its critical points:

Notice as $x\to\infty$, $h(x)\to\infty$, and as $x\to-\infty$ $h(x)\to-\infty$. So Theorem 8.3.3 doesn't exactly apply. Instead, let's consider the signs of $h'(x)$.

So, $h(x)$ increases until $x=-2$, then decreases. That means $h(x)$ has a local maximum at $x=-2$. The function decreases from $-2$ until $2$, after which is increases, so $h(x)$ has a local minimum at $x=2$. We compute $f(-2)=20$ and $f(2)=-12$.

One way to decide whether a critical point $x=c$ is a local extremum is to consider the first derivative. For example: if $f'(x)$ is negative for all $x$ just to the left of $c$, and positive for all $x$ just to the right of $c$, then $f(x)$ decreases up till $c$, then increases after $c$, so $f(x)$ has a local minimum at $c$.

$(-2,33)$ max, and $(2,-31)$ min

Since $h(x)$ is a polynomial, it has no singular points. We compute its critical points:

Notice as $x\to\infty$, $h(x)\to\infty$, and as $x\to-\infty$ $h(x)\to-\infty$. So Theorem 8.3.3 doesn't exactly apply. Instead, let's consider the signs of $h'(x)$.

So, $h(x)$ increases until $x=-2$, then decreases. That means $h(x)$ has a local maximum at $x=-2$. The function decreases from $-2$ until $2$, after which is increases, so $h(x)$ has a local minimum at $x=2$.

We compute $f(-2)=33$ and $f(2)=-31$.

Start with a formula for travel time from $P$ to $B$. You might want to assign a variable to the distance from $A$ where your buggy first reaches the road.

$Q$ should be $4\sqrt{3}$ kilometres from $A$

Suppose that $Q$ is a distance of $x$ from $A$. Then it is a distance of $18-x$ from $B$.

Using the Pythagorean Theorem, the distance from $P$ to $Q$ is $\sqrt{12^2+x^2}$ kilometres, and the buggy travels 15 kph over this off-road stretch. The travel time from $P$ to $Q$ is $\dfrac{\sqrt{12^2+x^2}}{15}$ hours.

The distance from $Q$ to $B$ is $18-x$ kilometres, and the dune buggy travels $30$ kph along this road. The travel time from $Q$ to $B$ is $\dfrac{18-x}{30}$ hours. So, the total travel time is

We wish to minimize this for $0\le x\le 18$. We will test all singular points, critical points, and endpoints to find which yields the smallest value of $f(x)$. Since there are no singular points, we begin by locating the critical points.

So the minimum travel times must be one of $f(0)$, $f(18)$, and $f\left(4\sqrt{3}\right)$.

So $Q$ should be $4\sqrt{3}$ km from $A$.

A box has three dimensions; make variables for them, and write the relations given in the problem in terms of these variables.

$10\times 30 \times 15$

Let $\ell,\ w$ and $h$ denote the length, width and height of the box respectively. We are told that $\ell w h =4500$ and that $\ell=3w$. Hence $h=\dfrac{4500}{\ell w}=\dfrac{4500}{3 w^2}=\dfrac{1500}{w^2}$. The surface area of the box is

As $w$ tends to zero or to infinity, the surface area approaches infinity. By Theorem 8.3.3 the minimum surface area must occur at a critical point of $w^2+\dfrac{2000}{w}$.

Therefore,

The dimensions of the box with minimum surface area are $10 \times 30 \times 15$.

Find a formula for the cost of the base, and another formula for the cost of the other sides. The total cost is the sum of these two formulas.

$2\times 2\times 6$

Let the length of the sides of the square base be $b$ metres and let the height be $h$ metres. The area of the base is $b^2$, the area of the top is $b^2$ and the area of each of the remaining four sides is $bh$ so the total cost is

Solving for $h$,

The volume is

This is the function we want to maximize. Since volume is never negative, the endpoints of the functions are the values of $b$ that make the volume 0. So, the maximum volume will not occur at an endpoint, it will occur at a critical point. The only critical point is $b=2$:

The desired dimensions are $2\times 2\times 6$.

The setup is this:

$X=Y=\sqrt{2}$

It suffices to consider $X$ and $Y$ such that the line $XY$ is tangent to the circle. Otherwise we could reduce the area of the triangle by, for example, holding $X$ fixed and reducing $Y$. So let $X$ and $Y$ be the $x$– and $y$–intercepts of the line tangent to the circle at $(\cos\theta,\sin\theta)$. Then $\dfrac{1}{X}=\cos\theta$ and $\dfrac{1}{Y}=\cos\left(\dfrac{\pi}{2}-\theta\right) =\sin\theta$. The area of the triangle is

This is a minimum when $\sin(2\theta)$ is a maximum. That is when $2\theta=\dfrac{\pi}{2}$. Hence $X=\dfrac{1}{\cos(\pi/4)}$ and $Y=\dfrac{1}{\sin(\pi/4)}$. That is, $X=Y=\sqrt{2}$.

Put the whole system on $xy$-axes, so that you can easily describe the pieces using $(x,y)$-coordinates.

The largest possible perimeter is $2\sqrt{5}R$ and the smallest possible perimeter is $2R$.

For ease of notation, we place the semicircle on a Cartesian plane with diameter along the $x$-axis and centre at the origin.

If $x$ is the point where the rectangle touches the diameter to the right of the $y$-axis, then $2x$ is the width of the rectangle. The origin and the two right corners of the rectangle form a right triangle with hypotenuse $R$, so by the Pythagorean Theorem, the upper right hand corner of the rectangle is at $\big(x,\sqrt{R^2-x^2}\big)$. The perimeter of the rectangle is given by the function:

So, this is what we optimize. The endpoints of the domain for this function are $x=0$ and $x=R$. To find the critical points, we differentiate:

Note that since our perimeter formula was defined to work only for $x$ in $[0,R]$, we neglect the negative square root, $-\dfrac{2}{\sqrt{5}}R$.

Now, we find the size of the perimeter at the critical point and the endpoints:

So, the largest possible perimeter is $2\sqrt{5}R$ and the smallest possible perimeter is $2R$.

Remark: as a check on the correctness of our formula for $P(x)$, when $x=0$ the rectangle degenerates to the line segment from $(0,0)$ to $(0,R)$. The perimeter of this “width zero rectangle” is $2R$, agreeing with $P(0)$. Similarly, when $x=R$ the rectangle degenerates to the line segment from $(R,0)$ to $(-R,0)$. The perimeter of this “width zero rectangle” is $4R$, agreeing with $P(R)$.

The surface area consists of two discs and a strip. Find the areas of these pieces.

The volume of a cylinder with radius $r$ and height $h$ is $\pi r^2 h$.

$\dfrac{A^{3/2}}{3\sqrt{6\pi}}$

Let the cylinder have radius $r$ and height $h$. If we imagine popping off the ends, they are two circular disks, each with surface area $\pi r^2$. Then we imagine unrolling the remaining tube. It has height $h$, and its other dimension is given by the circumference of the disks, which is $2\pi r$. Then the area of the “unrolled tube" is $2\pi rh$.

So, the surface area is $2\pi r^2+2\pi r h$. Since the area is given as $A$, we can solve for $h$:

Then we can write the volume as a function of the variable $r$ and the constant $A$:

This is the function we want to maximize. Let's find its critical points.

since negative values of $r$ don't make sense. At this critical point,

We should also check the volume of the cylinder at the endpoints of the function. Since $r \geq 0$, one endpoint is $r=0$. Since $h \geq 0$, and $r$ grows as $h$ shrinks, the other endpoint is whatever value of $r$ causes $h$ to be 0. We could find this value of $r$, but it's not strictly necessary: when $r=0$, the volume of the cylinder is zero, and when $h=0$, the volume of the cylinder is still zero. So, the maximum volume does not occur at the endpoints.

Therefore, the maximum volume is achieved at the critical point, where

Remark: as a check, $A$ has units $m^2$ and, because of the $A^{3/2}$, our answer has units $m^3$, which are the correct units for a volume.

If the circle has radius $r$, and the entire window has perimeter $P$, what is the height of the rectangle?

$\dfrac{P^2}{2(\pi+4)}$

Denote by $r$ the radius of the semicircle, and let $h$ be the height of the recangle.

Since the perimeter is required to be $P$, the height, $h$, of the rectangle must obey

So the area is

Finding all critical points:

Now we want to know what radius yields the maximum area. We notice that $A'(r)>0$ for $r<\dfrac{P}{\pi+4}$ and $A'(r)<0$ for $r>\dfrac{P}{\pi+4}$. So, $A(r)$ is increasing until the critical point, then decreasing after it. That means the global maximum occurs at the critical point, $r=\dfrac{P}{\pi+4}$. The maximum area is

Remark: another way to see that the global maximum occurs at the critical point is to compare the area at the critical point to the areas at the endpoints of the function. The smallest value of $r$ is 0, while the biggest is $\dfrac{P}{\pi+2}$ (when the shape is simply a half-circle). Comparing $A(0)$, $A\left(\dfrac{P}{\pi+2}\right)$, and $A\left(\dfrac{P}{\pi+4}\right)$ is somewhat laborious, but certainly possible.

(a) $x=\sqrt{\dfrac{A}{3p}}$, $y=\sqrt{\dfrac{Ap}{3}}$, and $z=\dfrac{\sqrt{Ap}}{\sqrt{3}(1+p)}$

(b) $p=1$

(The dimensions of the resulting baking pan are $x=y=\sqrt{ \dfrac{A}{3} }$ and $z=\dfrac{1}{2}\sqrt{ \dfrac{A}{3} }$.)

(a) The surface area of the pan is

and the volume of the pan is $xyz=px^2z$. Assuming that all $A\,{\rm cm}^2$ is used, we have the constraint

So

Using the product rule,

The derivative $V'(x)$ is 0 when $x=\sqrt{\dfrac{A}{3p}}$. The derivative is positive (i.e. $V(x)$ is increasing) for $x< \sqrt{\dfrac{A}{3p}}$ and is negative (i.e. $V(x)$ is decreasing) for $x>\sqrt{\dfrac{A}{3p}}$. So the pan of maximum volume has dimensions $x=\sqrt{\dfrac{A}{3p}}$, $y=p\sqrt{\dfrac{A}{3p}}=\sqrt{\dfrac{Ap}{3}}$ and $z=\dfrac{2A/3}{2(1+p)\sqrt{A/(3p)}}=\dfrac{\sqrt{Ap}}{\sqrt{3}(1+p)}$. (b) The volume of the pan from part (a) is

Since

the volume is increasing with $p$ for $p<1$ and decreasing with $p$ for $p>1$. So the maximum volume is achieved for $p=1$ (a square base).

Use logarithmic differentiation to find $f'(x)$.

(a) $x^x(1+\log x)$ (b) $x=\dfrac{1}{e}$ (c) local minimum

(a) We use logarithmic differentiation.

(b) Since $x>0$, $x^x>0$. Therefore,

(c) Since $x>0$, $x^x>0$. So, the sign of $f'(x)$ is the same as the sign of $1+\log x$.

For $x<\dfrac{1}{e}$, $\log x<-1$ and $f'(x)<0$. That is, $f(x)$ decreases as $x$ increases, when $x<\dfrac{1}{e}$. For $x>\dfrac{1}{e}$, $\log x>-1$ and $f'(x)>0$. That is, $f(x)$ increases as $x$ increases, when $x>\dfrac{1}{e}$. Hence $f(x)$ is a local minimum at $x=\dfrac{1}{e}$.

When you are finding the global extrema of a function, remember to check endpoints as well as critical points.

Maximum area: do not cut, make a circle and no square.
Minimum area: make a square out of a piece that is $\dfrac{4}{4+\pi}$ of the total length of the wire.

Call the length of the wire $L$ units and suppose that it is cut $\ell$ units from one end. Make the square from the piece of length $\ell$, and make the circle from the remaining piece of length $L-\ell$.

The square has perimeter $\ell$, so its side length is $\ell/4$ and its area is $\left(\dfrac{\ell}{4}\right)^2$. The circle has circumference $L-\ell$, so its radius is $\dfrac{L-\ell}{2\pi}$ and its area is $\pi\left(\dfrac{L-\ell}{2\pi}\right)^2=\dfrac{(L-\ell)^2}{4\pi}$.

The area enclosed by the shapes, when the square is made from a length of size $\ell$, is

We want to find the global max and min for this function, given the constraint $0 \leq \ell \leq L$, so we find its derivative:

Now, we find the critical point.

It seems obnoxious to evaluate $A\left(\frac{4L}{\pi+4}\right)$, and the problem doesn't ask for it–but we still have to figure out whether it is a global max or min.

When $\ell<\frac{4L}{\pi+4}$, $A'(\ell)<0$, and when $\ell>\frac{4L}{\pi+4}$, $A'(\ell)>0$. So, $A(\ell)$ is decreasing until $\ell=\frac{4L}{\pi+4}$, then increasing. That means our critical point $\ell=\frac{4L}{\pi+4}$ is a local minimum.

So, the minimum occurs at the only critical point, which is $\ell=\left(\dfrac{4}{4+\pi}\right)L$. This corresponds to $\dfrac{\ell}{L}=\dfrac{4}{4+\pi}$: the proportion of the wire that is cut is $\dfrac{4}{4+\pi}$.