최댓값·최솟값과 최적화

The radius should be about $3.84$cm and the height should be $2r=7.67$cm. No, this is not the size of the standard can.

The radius should be about $3.2$in and the height should be $2r=6.4$in. As the #10 is not a perfect cylinder (with extra material to aid in stacking, etc.), the dimensions are close enough to assume that minimizing surface area was a consideration.

The height and width should be $18$ and the length should be $36$, giving a volume of $11,664$in$^3$.

The dog should run about 13 feet along the shore before starting to swim.

The farmer will fence a rectangular area whose length $x$ and width $y$ are unknown. The total area of the rectangular area is composed of 4 equal area rectangles. There are two ways to do this as shown in the figure below (though only the one on the left explicitly follows the way the problem is framed).

We want to maximize the area of each subrectangular region, which is the same as maximizing the total area $A = xy$ of the entire rectangular region. Since $A$ has two variables, we can't yet directly apply calculus techniques to it. We use the information about the amount of fencing to relate $x$ and $y$ in order to write $A$ in terms of just one variable.

In the case of the fencing option on the left in the figure, the total amount of fencing that is needed to enclose the four grazing areas is $2x+5y$. We assume that all of the $7500$ feet of fencing will be used, so $2x+5y=7500$. Solving for $x$, $x = 3750 - 2.5y$. Substituting this expression for $x$ into the area formula allows us to write $A$ in terms of $y$ alone:

$$A(y) = (3750-2.5y)y = 3750y - 2.5y^2\,.$$ To enclose any area, we of course must have $y > 0$ and $x > 0$, and since $x = 3750 - 2.5y$, it follows that we need to have $2.5y < 3750$ so $y < 1500$. To find the maximum value of $A$ on this domain, we find the critical numbers of $A$. Since $A'(y) = 3750 - 5y$, we see that $A'(y) = 0$ when and only when $y = 750$. For $0 < y < 750$, we see that $A'(y) > 0$ and for $750 < y < 1500$ we have $A'(y) < 0$. So $A(750) = 1406250$ square feet is the absolute maximum value of $A$. The maximum possible area that each of the four pens can enclose is one quarter of the total, or $351562.5$ square feet.

In the case of the fencing option on the right in the figure, the total amount of fencing that is needed to enclose the four grazing areas is $3x+3y$. We assume that all of the $7500$ feet of fencing will be used, so $3x + 3y = 7500$. Solving for $x$ and substituting into $A$, we find $x = 2500 - y$ and

$$A(y) = (2500-y)y = 2500y - y^2\,.$$ To enclose any area, we of course must have $y > 0$ and $x > 0$, and since $x = 2500 - y$, it follows that we need to have $y < 2500$. To find the maximum value of $A$ on this domain, we find the critical numbers of $A$. Since $A'(y) = 2500 - 2y$, we see that $A'(y) = 0$ when and only when $y = 1250$. For $0 < y < 1250$, we see that $A'(y) > 0$ and for $1250 < y < 2500$ we have $A'(y) < 0$. So $A(1250) = 1406250$ square feet is the absolute maximum value of $A$. The maximum possible area that each of the four pens can enclose is one quarter of the total, or $390625$ square feet.

So the maximum possible area that each of the four pens can enclose is 390625 square feet, and we see that the righthand alternative is even more optimal than the left one.

We let the length of the cylindrical part of the tank be $h$ with the radius of the cylinder given by $r$. It follows that the radius of the hemispherical ends of the tank is also $r$. We first identify formulas for both the volume and surface area of the tank by looking at the cylindrical and hemispherical pieces.

The cylindrical part of the tank has volume $V_c = \pi r^2 h$, and the surface area of its sides is $A_c = 2\pi r h$ (which is the circumference of the cylinder's base times the cylinder's height -- think about how a cylinder unfurls to form a rectangle). The two hemispherical ends together constitute a full sphere, and the volume of a sphere is $V_s = \frac{4}{3} \pi r^3$, while the surface area of a sphere is $A_s = 4 \pi r^2$.

The overall volume of the tank is therefore

$$V = V_c + V_s = \pi r^2 h + \frac{4}{3} \pi r^3\,,$$ and its total surface area is

$$A = A_c + A_s = 2\pi r h + 4 \pi r^2\,.$$ Moreover, taking into account the different costs of material for the cylindrical sides and hemispherical ends, we find that the total cost of the tank is

$$C = 2(2\pi r h) + 5(4 \pi r^2) = 4 \pi r h + 20 \pi r^2\,.$$

Our goal is to minimize the total cost of the tank given that we want it to hold 1000 cubic meters of volume. From this constraint, we have the equation $1000 = \pi r^2 h + \frac{4}{3} \pi r^3$, which we can solve for $h$ to get $\pi r^2 h = 1000 - \frac{4}{3} \pi r^3$, so

$$h = \frac{1000 - \frac{4}{3} \pi r^3}{\pi r^2}\,.$$ This allows us to now write cost as a function of the single variable $r$ by substituting the expression for $h$ into the earlier formula we found for cost. In particular,

$$C(r) = 4 \pi r \frac{1000 - \frac{4}{3} \pi r^3}{\pi r^2} + 20 \pi r^2 = 4 \frac{1000 - \frac{4}{3} \pi r^3}{r} + 20\pi r^2$$

Simplifying further, we have found that

$$C(r) = \frac{4000}{r} - \frac{16}{3}\pi r^2 + 20 \pi r^2 = \frac{4000}{r} + \frac{44 \pi}{3}r^2\,.$$ Based on the physical constraints of the problem, we obviously will only consider values of $r > 0$ and $h > 0$. Since $h = \frac{1000 - \frac{4}{3} \pi r^3}{\pi r^2}$, we must have

$$\frac{4}{3} \pi r^3 < 1000$$

so $r^3 < \frac{3}{4 \pi} \cdot 1000 = 750$, and thus $r < \sqrt[3]{\frac{750}{\pi}} \approx 6.2035$. We now work to minimize $C(r)$ on the interval $0 < r < \sqrt[3]{\frac{750}{\pi}}$.

Taking the derivative of $C$, we find that $C'(r) = -\frac{4000}{r^2} + \frac{88 \pi}{3}r$. Setting $C'(r) = 0$, it follows that

$$\frac{4000}{r^2} = \frac{88 \pi}{3}r$$

and thus

$$4000 \cdot \frac{3}{88 \pi} = r^3\,.$$ Therefore, $r^3 = \frac{1500}{11 \pi}$, so $r = \sqrt[3]{\frac{1500}{11 \pi}} \approx 3.5144$ is the only critical number of $C$; observe that it lies within the interval $0 < r < \sqrt[3]{\frac{750}{\pi}} \approx 6.2035$.

If we plot the derivative function, $y = C'(r)$, on the interval $0 < r < \sqrt[3]{\frac{750}{\pi}}$, we see that for $0 < r < 3.5144$, $C'(r) < 0$, and for $3.5144 < r < 6.2035$, $C'(r) > 0$, which shows that $C$ is decreasing before the critical number and increasing after the critical number, and hence $C$ has a global minimum of $C(\sqrt[3]{\frac{1500}{11 \pi}}) \approx 1707.27$, so $1707.27 is the minimum cost to build the desired tank under the given constraints.