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A cylindrical container is to be constructed to hold $32\pi$ cm3 of flammable material. The cost per cm2 of constructing the top is three times that of the sides and bottom, since the top must contain a safety valve. What dimensions will minimize cost of construction?
The surface area is given by the sum of the areas of the top, bottom, and sides:
$$A = \pi r^2 + \pi r^2 + 2\pi rh$$The cost per unit area of the top is three times that of the sides and bottom, so we have a cost function expressible as
$$C = 3\pi r^2 + \pi r^2 + 2\pi rh = 4\pi r^2 + 2\pi rh$$The given volume tells us $\pi r^2 h = 32\pi$, which we can solve for $h$ to find $h = 32r^{-2}$. This can be substituted into the above equation to get the cost in terms of a single variable $r$:
$$C(r) = 4\pi r^2 + 64 \pi r^{-1}$$Finding $C'(r) = 8\pi r - 64\pi r^{-2}$ leads to an absolute minimum when $r=2$ cm and $h = 8$ cm.
Find two positive numbers whose product is $1$ such that the sum of the first number plus the square of the second is a minimum.
If $x,y$ are the two numbers, they're related by $xy=1$ and we want to minimize $x+y^2=x+1/x^2$ over $(0,\infty)$. The min is at $x=\sqrt[3]{2}$ (the limits of the endpoints both give $+\infty$).
Find the point of $y=\sqrt{x}$ closest to $(c,0)$ if $c \ge \frac{1}{2}$ and if $c \lt \frac{1}{2}$. (Hint: minimize the square of the distance.)
We're trying to minimize the function $(x-c)^2 - (0-\sqrt{x})^2$ over $[0,\infty)$ (the domain of $\sqrt{x}$). The function has a min at $x=c-\frac{1}{2}$, but this point must be in our domain, so if $c \gt 1/2$ the only minimum is the point $\left(c-\frac{1}{2},\sqrt{c-\frac{1}{2}}\right)$; if $c=\frac{1}{2}$ we have two minima at $x=c-\frac{1}{2}$ and $x=0$ (in both cases the distance squared is $1/4$); finally if $c \lt 1/2$ the minimum would be negative, so the closest point on the function is the origin.
Farmer Brown is standing in a field, and wants to get home as soon as possible. Along the field runs a straight road, that leads directly to his house. Farmer Brown is 1 mi from the road and 5/3 mi from his house, and can walk through the field at 2 mi/h and on the road at 4 mi/h. What’s the soonest that Farmer Brown can be home?
Let $A$ be the point on the road closest to Farmer Brown, and $C$ be his house. A simple pythagorean theorem computation shows that the distance from $A$ to $C$ is $4/3$. Suppose that Farmer Brown crosses through the field until a point $B$ on the road that’s $x$ miles from $A$, and walks on the road the remaining time. Since time is distance over speed, this path takes a time of
$$\frac{\sqrt{1+x^2}}{2} + \frac{(\frac{4}{3} - x)}{4} \textrm{ hours}$$Minimizing this over the interval $[0,4/3]$ we get that the min is at $x = 1/\sqrt{3}$ mi, where it takes Farmer Brown $\frac{\sqrt{3}}{4}+\frac{1}{3}$ hours to get home, as opposed to the value $5/6$ at both endpoints.
A rectangle is inscribed between the $x$-axis and the parabola $y=a^2-x^2$, one side of the rectangle lying on the $x$-axis. Find the rectangle of this kind with the largest area.
If you draw the problem you quickly realize that, by symmetry, if $l$ is half the length of the side on the $x$-axis (so the point ($l$, 0) is a vertex of the rectangle), the area is given by $2l(a^2-l^2)$, with $l$ in $[0,a]$. You can easily maximize this with the value $l = a/\sqrt{3}$, in which case the height is $\frac{2}{3}a^2$ and the area is $\frac{4a^3}{3\sqrt{3}}$.
An open rectangular box is formed by taking a square sheet of cardboard and removing identical squares from each of the corners, then folding it up. What’s the largest volume the box can reach?
If the big square has side $L$ (a constant), and the small square has side $x$, the resulting box has height $x$ and a square base of side $L-2x$, so its volume is $x(L-2x)^2$, which we need to maximize on $[0,1/2]$. Taking the derivative and solving the quadratic equation yields the max $x=L/6$, where the volume is $V = \frac{2}{27}L^3$.
A silo is in the shape of a cylinder (with no base) topped by a hemisphere. The cost per square foot of constructing the hemisphere is twice as big as the cost per square foot of constructing the cylinder. If the volume of the silo is fixed, what are the dimensions of the silo whose construction cost is minimal?
If $h$ is the height of the cylinder and $r$ is the radius of both the cylinder and the hemisphere (they must match), then the volume is $v = \pi r^2 h + \frac{2}{3} \pi r$ and the cost (which is the cost per square foot times the area) is given by $C = 2 \cdot 2\pi r^2 + 1 \cdot (2\pi r h + \pi r^2)$. Solving the volume for $h$ and plugging into $C$, we want to minimize $C(r) = \frac{11}{3} \pi r^2 + 2V/r$ over $(0,(\frac{3}{2}V)^{1/3})$. The min is at $r = \sqrt[3]{3V/(8\pi)}$. Geometrically, this solution gives $h = 2r$, so there's some nice symmetry.).
A box with no top has base length twice its width. Find the box of maximum volume made from 24 square feet of material.
Dimensions: $2 \times 4 \times \frac{4}{3}$ (in feet)
Let $x$ and $y$ be legs of a right triangle whose hypotenuse is 2. For what values of $x$ and $y$ is $2x+y^2$ a maximum?
$x = 1, y = \sqrt{3}$
A rectangle box is to have a square base and top and is to hold 1000 in3 of material. What are the dimensions of the box if it is to have a minimum surface area?
$10$ inches $\times$ $10$ inches
A window in the shape of a rectangle capped by a semicircle is to be surrounded by $p$ inches of metal border. Find the radius of a semicircular part if the total area of the window is to be a maximum.
$\displaystyle{r = \frac{p}{4+\pi}}$
A rectangular garden is to be laid out along a neighbor's property line and is to contain 432 yds2. If the neighbor pays for half of the dividing fence, what are the dimensions so that the cost to the owner is minimized?
$18$ yds $\times$ $24$ yds
A rectangular box with top has base length three times its width. Find the dimensions of such a box of maximum volume that can be constructed from 200 in2 of material.
$10$ inches $\times$ $\frac{10}{3}$ inches $\times$ $5$ inches
A rectangle has an area of $8$ in2. Find the dimensions so that the distance from one corner to the midpoint of the opposite side is minimized.
($x=2$ in, $y=4$ in)
A museum display case in the shape of a rectangular box with square base and a volume of 54 ft3 is to be built. The front, back and sides are to be made of glass, which costs $10$ dollars per square foot; the top and bottom are to be made of fine wood, which costs $20$ dollars per square foot.
Find two positive numbers whose product is 100 and whose sum is a minimum.
Find the volume of the largest right circular cylinder that can be inscribed in a sphere of radius $r$.
Two hallways of widths a feet and b feet intersect at right angles. What is the longest ladder that can be carried horizontally around the corner?
A metal trough is formed from a metal sheet by folding the sheet lengthwise down the middle (so that the cross-section of the trough is a triangle) and capping both ends. At what angle should the sheet be folded to form the trough of maximum volume?