Since we have expressed the cost as a function of a single variable, the problem becomes a“calculus problem”: Find the minimum of the function $C$ on the interval $(0,\infty )$.

Although the function $C$ is continuous on its domain, its domain is an open, unbounded interval and the Extreme Value Theorem does not apply; there is no guarantee that the minimum even exists. But we don’t give up! Our strategy will be to search for a critical point and check if the minimum is attained there. By solving the equation $$C'(r)=-2cV/r^2+4Nc\pi \answer [given]{r} =0$$ we find that the point where $r=\sqrt [3]{V/(2N\pi )}$ is the only critical point of $C$. Since $C''(r)=4cV/r^3+4Nc\pi$ is positive when $r$ is positive, there is a local minimum at the critical point, and hence a global minimum since there is only one critical point.

Another way to reach this conclusion is to examine the sign of the derivative $C'(r)$ on the interval $(0,\infty )$: $$C'(r)=-2cV/r^2+4Nc\pi r=\frac {4Nc\pi (r^3-\frac {V}{2N\pi })}{r^2}.$$ From this factorization it is clear that $C'(r)<0$ on $(0,\sqrt [3]{V/(2N\pi )})$, and $C'(r)>0$ on $(\sqrt [3]{V/(2N\pi )}, \infty )$. This means that the function $C$ is decreasing on $(0,\sqrt [3]{V/(2N\pi )})$ and increasing on $(\sqrt [3]{V/(2N\pi )}, \infty )$, and, therefore, has a global minimum at the critical point.

Finally, since $h=V/(\pi r^2)$,

so the minimum cost occurs when the height $h$ is $2N$ times the radius. If, for example, there is no difference in the cost of materials, the height is twice the radius.

We want to know the maximum value of the function $P$ on the interval $[0,1.5]$. The Extreme Value Theorem now guarantees that the maximum is attained, since the function $P$ is continuous on the closed interval, $[0,1.5]$. We have to find the critical points: $$P'(x)=0,$$ or $$-20000x+25000=0,$$ which implies that the only critical point occurs at $x=1.25$. Now we have to evaluate the function at the critical point and both endpoints, and compare the values:

$P(0)=-12000$, $P(1.25)=3625$, and $P(1.5)=3000$. Note that $P(1.25)$ is the maximum of these. Thus the maximum profit is $$3625$, attained when we set the price at $$1.25$ and sell $7500$ items.

Alternately, we could apply the Second Derivative Test at the critical point and conclude that there must be a local maximum there, since $P''(x)=-20000<0$. Since this is the only critical point, it must be a global maximum as well. We can confirm our results by looking at the graph of $y=P(x)$: $$\graph [xmin=0,xmax=3,ymin=-4000,ymax=6000]{y=-10000x^2+25000x-12000}$$

Notice that the function we want to maximize, $\pi r^2h/3$, depends on two variables. Our next step is to find the relationship between $r$ and $h$ and use it to solve for one of the variables in terms of the other, so as to have a function of only one variable to maximize. In this problem, the condition is apparent in the figure below.

Apply the Pythagorean Theorem to the right triangle in the figure above to obtain

$$r^2 + (h-R)^2=R^2.$$ Solve for $r^2$, since $r^2$ is found in the formula for the volume of the cone, and find that $$r^2=R^2-(h-R)^2=2hR-h^2.$$ Substitute this into the formula for the volume of the cone to find that $$V_{\text {c}}(h)=\pi (R^2-(h-R)^2)h/3.$$ By simplifying, we were able to express the volume of the cone as a function of a single variable, $h$, $$V_{\text {c}}(h)=-{\pi \over 3}h^3+{2\over 3}\pi h^2R.$$

We want to maximize the function $V_{\text {c}}$ on the interval $[0,2R]$. Since this is a continuous function on a closed interval, the maximum is attained either at a critical point or an end point. By solving the equation $$V_{\text {c}}'(h)=-\pi h^2+(4/3)\pi h R=0,$$ we find that the function has at $h=4R/3$ its only critical point, since $h=0$ is an end point. We evaluate the function at all the critical and end points, $$V_{\text {c}}(0)=V_{\text {c}}(2R)=0,$$

$$V_{\text {c}}(4R/3)=(32/81)\pi R^3.$$

The maximum is obviously attained at $h=4R/3$ . Since the volume of the sphere is $(4/3)\pi R^3$, the fraction of the sphere occupied by the cone is $$\frac {(32/81)\pi R^3}{(4/3)\pi R^3}=\frac {8}{27}\approx 30\%.$$

The optimal solution likely has the line being run along the ground for a while, then underwater, as the figure implies. We need to label our unknown distances: the distance run along the ground and the distance run underwater. Recognizing that the underwater distance can be measured as the hypotenuse of a right triangle, we can label our figure as follows

Evaluating $c(x)$ at $x=416.67$ gives a cost of about $\$370000$. The distance the power line is laid along land is $5000-416.67 = 4583.33\unit {ft}$ and the underwater distance is $\sqrt {416.67^2+1000^2} \approx 1083\unit {ft}$. We can confirm our results by looking at the graph of $y=c(x)$: $$\graph [xmin=0, xmax=5000, ymin=0, ymax=800000]{y=50(5000-x)+130\sqrt {x^2+1000000}}$$

You travel the distance from $D$ to $B$ at speed $v$, and then the distance from $B$ to $A$ at speed $w$. The distance from $D$ to $B$ is $a-x$. By the Pythagorean theorem, the distance from $B$ to $A$ is $$\sqrt {x^2+b^2}.$$ Hence the total time for the trip is $$T(x)=\frac {a-x}{v}+\frac {\sqrt {x^2+b^2}}{w}.$$ We want to find the minimum value of $T$ when $x$ is between 0 and $a$. As usual we set $T'(x)=0$ and solve for $x$. Write $$T'(x)=\answer [given]{-1/v+\frac {x}{w\sqrt {x^2+b^2}}} =0.$$ We find that $$x=\frac {wb}{\sqrt {v^2-w^2}}$$ Notice that $a$ does not appear in the last expression, but $a$ is not irrelevant, since we are interested only in critical values that are in $[0,a]$, and $wb/\sqrt {v^2-w^2}$ is either in this interval or not. If it is, we can use the second derivative to test it: $$T''(x) = \answer [given]{\frac {b^2}{(x^2+b^2)^{3/2}w}}.$$ Since this is always positive there is a local minimum at the critical point, and so it is a global minimum as well.

If the critical value is not in $[0,a]$ it is larger than $a$. In this case the minimum must occur at one of the endpoints. We can compute

but it is difficult to determine which of these is smaller by direct comparison. If, as is likely in practice, we know the values of $v$, $w$, $a$, and $b$, then it is easy to determine this. With a little cleverness, however, we can determine the minimum in general. We have seen that $T''(x)$ is always positive, so the derivative $T'(x)$ is always increasing. We know that at $wb/\sqrt {v^2-w^2}$ the derivative is zero, so for values of $x$ less than that critical value, the derivative is negative. This means that $T(0)>T(a)$, so the minimum occurs when $x=a$.

So the upshot is this: If you start farther away from $C$ than $wb/\sqrt {v^2-w^2}$ then you always want to cut across the sand when you are a distance $wb/\sqrt {v^2-w^2}$ from point $C$. If you start closer than this to $C$, you should cut directly across the sand.