Corrected exercises: optimization problems (MATH 140)

a) Counting the fence on the figure: three pieces of length $x$ and one of length $y$, so the constraint is $3x + y = 600$, that is $y = 600 - 3x$. The objective is the total area $A = xy$, and substituting the constraint, $A(x) = x(600 - 3x) = 600x - 3x^2$. The domain comes from the geometry, not from the formula: $x \ge 0$ and $y \ge 0$, and $y = 600 - 3x \ge 0$ gives $x \le 200$. So $x \in [0, 200]$. The endpoints are degenerate pastures of area $0$ (no depth, or all the fence used across the river), and including them costs nothing: it makes the domain a CLOSED interval, on which the continuous polynomial $A$ is guaranteed an absolute maximum by the Extreme Value Theorem. That is the licence for the closed interval method.

b) $A'(x) = 600 - 6x$, which is $0$ for $x = 100$, inside $[0, 200]$. There is no point where $A'$ fails to exist. Closed interval method: compare the values at the critical number and at BOTH endpoints: $A(0) = 0$, $A(200) = 0$, $A(100) = 100 \times 300 = 30\,000$. The absolute maximum is $30\,000$ m$^2$, reached for $x = 100$ m and $y = 600 - 300 = 300$ m. The answer the question asks for is the DIMENSIONS and the AREA: a pasture $300$ m along the river and $100$ m deep, split into two pens of $100$ m by $150$ m, total area $30\,000$ m$^2$. Writing only $x = 100$ answers a question nobody asked.

c) The domain is now $[0, 80]$. The only critical number, $100$, is OUTSIDE it and is not a candidate. On $[0, 80]$, $A'(x) = 600 - 6x \ge 600 - 480 = 120 > 0$, so $A$ is increasing and its maximum is at the right endpoint: $x = 80$ m, $y = 600 - 240 = 360$ m, $A = 80 \times 360 = 28\,800$ m$^2$. The closed interval method gives the same answer: $A(0) = 0 < A(80)$. Reporting the unconstrained answer $x = 100$ here designs a pasture that does not fit on the land, and costs the whole question. The optimum of a constrained problem is often ON the boundary, with a derivative that is not $0$ there.

d) $A''(100) = -6 < 0$ is the second derivative test: it proves that $x = 100$ is a LOCAL maximum, that is, larger than the nearby values, and nothing about far away values. Two arguments make it global here, and one of them must be written. Either the closed interval method of b), which compares with the endpoints; or the fact that $x = 100$ is the ONLY critical number on the interval $[0, 200]$ and is a local maximum, so $A$ increases before it and decreases after it, and no other value can beat it. A local maximum with no justification of this kind is worth part of the marks only, because the next exercise with a larger endpoint will make the same sentence false.

e) Now $(n + 1)x + y = L$, so $A(x) = x(L - (n+1)x) = Lx - (n+1)x^2$ on $\left[0, \frac{L}{n+1}\right]$. $A'(x) = L - 2(n+1)x = 0$ gives $x = \frac{L}{2(n+1)}$, inside the interval, and $A$ vanishes at both endpoints, so this is the absolute maximum. Then $y = L - (n+1)\frac{L}{2(n+1)} = \frac{L}{2}$. Check with b): $L = 600$, $n = 2$ gives $x = 100$ and $y = 300$. Whatever the number of pens, half the fence runs along the river side: a result worth remembering to check a numerical answer in two seconds.

a) Removing a square of side $x$ at each end of a side shortens it by $2x$: the base is $(16 - 2x)$ by $(10 - 2x)$, and the folded flaps give the height $x$. So $V(x) = x(16 - 2x)(10 - 2x) = 4x(8 - x)(5 - x) = 4x(40 - 13x + x^2) = 4x^3 - 52x^2 + 160x$. Domain: every dimension must be $\ge 0$, so $x \ge 0$, $x \le 8$ and $x \le 5$. The SHORT side decides: $x \in [0, 5]$. The endpoints are flat boxes of volume $0$; they are included to make the interval closed.

b) $V'(x) = 12x^2 - 104x + 160 = 4(3x^2 - 26x + 40) = 4(3x - 20)(x - 2)$, as expanding $(3x - 20)(x - 2) = 3x^2 - 26x + 40$ confirms. The critical numbers are $x = 2$ and $x = \frac{20}{3}$. Since $\frac{20}{3} > 5$, it is NOT in the domain: cutting squares of side $\frac{20}{3} \approx 6.7$ cm from a sheet $10$ cm wide is impossible, the two cuts would overlap. The formula $V(x)$ knows nothing about cardboard; only the domain does. The candidate $x = 2$ is kept.

c) Closed interval method on $[0, 5]$: $V(0) = 0$, $V(5) = 5 \times 6 \times 0 = 0$, $V(2) = 2 \times 12 \times 6 = 144$. The largest volume is $144$ cm$^3$, for the box of base $12$ cm by $6$ cm and height $2$ cm. Check that $V$ is continuous on $[0, 5]$ (a polynomial) before invoking the method: that sentence is the licence, and it is one line. A quick plausibility check: $V(1) = 1 \times 14 \times 8 = 112$ and $V(3) = 3 \times 10 \times 4 = 120$, both below $144$.

d) $V\left(\frac{20}{3}\right) = \frac{20}{3} \cdot \frac{8}{3} \cdot \left(-\frac{10}{3}\right) = -\frac{1600}{27}$, a NEGATIVE volume, which says it all: the point is outside the domain and describes no box. The second derivative test was applied to a point that should never have been tested. The minimum volume on $[0, 5]$ is $0$, at the two endpoints, and it is a degenerate box. The general rule: reject a critical number outside the domain BEFORE any test, and say why in words.

e) $V(x) = x(a - 2x)^2$ on $\left[0, \frac{a}{2}\right]$. By the product and chain rules, $V'(x) = (a - 2x)^2 + x \cdot 2(a - 2x)(-2) = (a - 2x)\left[(a - 2x) - 4x\right] = (a - 2x)(a - 6x)$. Critical numbers: $x = \frac{a}{2}$ (an endpoint, $V = 0$) and $x = \frac{a}{6}$. With $V(0) = V\left(\frac{a}{2}\right) = 0$, the maximum is $V\left(\frac{a}{6}\right) = \frac{a}{6}\left(\frac{2a}{3}\right)^2 = \frac{a}{6} \cdot \frac{4a^2}{9} = \frac{2a^3}{27}$. For $a = 18$: $x = 3$ cm and $V = \frac{2 \times 5832}{27} = 432$ cm$^3$, which checks directly: $3 \times 12 \times 12 = 432$. Factoring $V'$ by the common factor $(a - 2x)$ instead of expanding is the gesture that keeps the numbers small.

a) The net has two disks of area $\pi r^2$ and a rectangle of area $2\pi r \cdot h$, so $A = 2\pi r^2 + 2\pi rh$. The constraint is the volume: $\pi r^2 h = 250\pi$, so $h = \frac{250}{r^2}$. Substituting, $A(r) = 2\pi r^2 + 2\pi r \cdot \frac{250}{r^2} = 2\pi r^2 + \frac{500\pi}{r}$. Solve the constraint for the variable that is easy to isolate, here $h$, never for $r$, which would bring a square root.

b) Any $r > 0$ gives a can, with $h = \frac{250}{r^2} > 0$, so the domain is $(0, \infty)$: open at $0$ and unbounded. As $r \to 0^+$, $\frac{500\pi}{r} \to \infty$ (a tall thin needle with an enormous side); as $r \to \infty$, $2\pi r^2 \to \infty$ (a huge flat coin). So $A(r) \to \infty$ at both ends. There is no endpoint at which to evaluate $A$, and the Extreme Value Theorem, which requires a closed interval, does not apply: nothing guarantees a minimum in advance. The justification must come from the derivative on the WHOLE domain.

c) $A'(r) = 4\pi r - \frac{500\pi}{r^2} = \frac{4\pi(r^3 - 125)}{r^2}$. On $(0, \infty)$ the denominator is positive, so the sign of $A'$ is that of $r^3 - 125$: negative for $0 < r < 5$, zero at $r = 5$, positive for $r > 5$. Hence $A$ is decreasing on $(0, 5]$ and increasing on $[5, \infty)$: $A(5)$ is smaller than EVERY other value, the first derivative test for absolute extreme values. (Equivalently, $A''(r) = 4\pi + \frac{1000\pi}{r^3} > 0$ on the whole domain, so the only critical number is the absolute minimum.) The can: $r = 5$ cm, $h = \frac{250}{25} = 10$ cm, and $A(5) = 50\pi + 100\pi = 150\pi$ cm$^2$. Answering the minimum is $5$ confuses the radius with the area: the question asked for dimensions and area.

d) The cost is proportional to $2 \cdot 2\pi r^2 + 2\pi rh$ (each disk counts twice), so we minimize $C(r) = 4\pi r^2 + \frac{500\pi}{r}$ on $(0, \infty)$. $C'(r) = 8\pi r - \frac{500\pi}{r^2} = \frac{4\pi(2r^3 - 125)}{r^2}$, negative then positive around $r^3 = \frac{125}{2}$, so the absolute minimum is at $r = \frac{5}{\sqrt[3]{2}} = \frac{5\sqrt[3]{4}}{2}$ cm. For the shape, use the constraint instead of computing $h$: $\frac{h}{r} = \frac{250}{r^3} = \frac{250}{125/2} = 4$. The cheap can is twice as tall, relative to its radius, as the can of c): a dearer lid pushes the design toward a smaller lid.

e) With $h = \frac{V}{\pi r^2}$, $A(r) = 2\pi r^2 + \frac{2V}{r}$ and $A'(r) = 4\pi r - \frac{2V}{r^2} = \frac{4\pi r^3 - 2V}{r^2}$, negative before and positive after $r^3 = \frac{V}{2\pi}$: the absolute minimum. Then $\frac{h}{r} = \frac{V}{\pi r^3} = \frac{V}{\pi} \cdot \frac{2\pi}{V} = 2$, so $h = 2r$: the height equals the diameter. c) is the case $V = 250\pi$: $h = 10 = 2 \times 5$. Real cans are taller than this because the lid is thicker metal and the seams cost too, which is exactly the effect measured in d).

a) $D(x) = (x - 3)^2 + (\sqrt x - 0)^2 = x^2 - 6x + 9 + x = x^2 - 5x + 9$, on the domain $[0, \infty)$ of the curve. $D'(x) = 2x - 5$, zero at $x = \frac{5}{2}$, inside the domain. The domain is not closed and bounded, so justify on the WHOLE domain: $D' < 0$ on $\left[0, \frac{5}{2}\right)$ and $D' > 0$ on $\left(\frac{5}{2}, \infty\right)$, so $D$ decreases then increases and $D\left(\frac{5}{2}\right)$ is its absolute minimum. $D\left(\frac{5}{2}\right) = \frac{25}{4} - \frac{25}{2} + 9 = \frac{25 - 50 + 36}{4} = \frac{11}{4}$. The closest point is $P\left(\frac{5}{2}, \sqrt{\frac{5}{2}}\right) = \left(\frac{5}{2}, \frac{\sqrt{10}}{2}\right)$, and the least distance is $\sqrt{\frac{11}{4}} = \frac{\sqrt{11}}{2}$. The endpoint $x = 0$ gives $D = 9$, larger, as it must.

b) $D(x) = \left(x - \frac{1}{4}\right)^2 + x = x^2 + \frac{x}{2} + \frac{1}{16}$ and $D'(x) = 2x + \frac{1}{2}$. The root $x = -\frac{1}{4}$ is NOT in the domain $[0, \infty)$: the curve $y = \sqrt x$ has no point there. On the domain, $D'(x) \ge \frac{1}{2} > 0$, so $D$ is increasing and its minimum is at the endpoint $x = 0$: the closest point is the origin, at distance $\sqrt{D(0)} = \sqrt{\frac{1}{16}} = \frac{1}{4}$. The student's equation had a solution and the problem still had its answer elsewhere, at an endpoint where $D' \ne 0$.

c) $D(x) = (x - a)^2 + x$, $D'(x) = 2x - 2a + 1$, zero at $x = a - \frac{1}{2}$. If $a \ge \frac{1}{2}$, this is in $[0, \infty)$, $D'$ changes from negative to positive there, and the closest point has $x = a - \frac{1}{2}$ (for $a = 3$, $x = \frac{5}{2}$, as in a)). If $0 < a < \frac{1}{2}$, the root is negative, $D' > 0$ on the domain, and the closest point is the origin (b) is $a = \frac{1}{4}$). Same method, two answers, and the switch happens exactly when the critical number leaves the domain.

d) On the branch, $y = \frac{4}{x}$, so $D(x) = x^2 + \frac{16}{x^2}$ on $(0, \infty)$, an OPEN domain. $D'(x) = 2x - \frac{32}{x^3} = \frac{2(x^4 - 16)}{x^3}$, negative for $0 < x < 2$ and positive for $x > 2$. So $x = 2$ gives the absolute minimum: the point $(2, 2)$, with $D(2) = 4 + 4 = 8$ and distance $\sqrt 8 = 2\sqrt 2$. The symmetry of the curve about $y = x$ predicted the point; the sign of $D'$ is what proves it. Note that $D \to \infty$ at both ends of the domain, as for the can of Exercise 3.

e) At $x = \frac{5}{2}$ the tangent slope is $\frac{1}{2\sqrt x} = \frac{1}{2\sqrt{5/2}}$. The slope of $AP$ is $\frac{\sqrt{5/2} - 0}{\frac{5}{2} - 3} = -2\sqrt{\frac{5}{2}}$. Their product is $-\frac{2\sqrt{5/2}}{2\sqrt{5/2}} = -1$: the shortest segment from a point to a smooth curve meets it at a right angle, a two second check of any closest point answer. As for $\frac{11}{4}$: it is the minimum of $D$, the SQUARE of the distance. The trick of minimizing $D$ changes the function, not the point; the value asked must be brought back through the square root.

a) The rectangle is $2x$ by $y = \sqrt{25 - x^2}$, so $A(x) = 2x\sqrt{25 - x^2}$. The corner must lie on the arc in the first quadrant: $0 \le x \le 5$. At $x = 0$ the rectangle is a vertical segment, at $x = 5$ a horizontal one, both of area $0$. $A$ is continuous on the closed interval $[0, 5]$.

b) Product rule, and chain rule with inner function $25 - x^2$: $A'(x) = 2\sqrt{25 - x^2} + 2x \cdot \frac{-2x}{2\sqrt{25 - x^2}} = \frac{2(25 - x^2) - 2x^2}{\sqrt{25 - x^2}} = \frac{50 - 4x^2}{\sqrt{25 - x^2}}$ for $0 \le x < 5$. $A'(x) = 0$ gives $x^2 = \frac{25}{2}$, $x = \frac{5}{\sqrt 2}$ (the negative root is outside the domain). $A'$ does not exist at $x = 5$: that point is a critical number in the wide sense, but it is an endpoint, already on the list of the closed interval method. Values: $A(0) = 0$, $A(5) = 0$, $A\left(\frac{5}{\sqrt 2}\right) = 2 \cdot \frac{5}{\sqrt 2} \cdot \sqrt{\frac{25}{2}} = 2 \cdot \frac{25}{2} = 25$. The largest rectangle has base $2x = 5\sqrt 2$, height $\frac{5}{\sqrt 2} = \frac{5\sqrt 2}{2}$ and area $25$, the square of the radius.

c) $A(x) \ge 0$ on $[0, 5]$, and $t \mapsto t^2$ is increasing on $[0, \infty)$, so $A$ and $A^2$ are largest at the same $x$. With $u = x^2 \in [0, 25]$: $A^2 = 4x^2(25 - x^2) = 4(25u - u^2)$, a downward parabola in $u$ with vertex $u = \frac{25}{2}$, inside $[0, 25]$. So $x = \frac{5}{\sqrt 2}$ again, with no square root to differentiate. The same trick as the distance in Exercise 4, with the same warning: the maximum of $A^2$ is $625$, the area is $\sqrt{625} = 25$. It is legitimate only because $A \ge 0$: squaring a quantity that changes sign can move its maximum.

d) The perimeter is $P(x) = 2(2x) + 2y = 4x + 2\sqrt{25 - x^2}$ on $[0, 5]$. $P'(x) = 4 - \frac{2x}{\sqrt{25 - x^2}}$, zero when $2\sqrt{25 - x^2} = x$. Both sides are $\ge 0$ on the domain, so squaring is safe: $4(25 - x^2) = x^2$, $x^2 = 20$, $x = 2\sqrt 5$, and then $y = \sqrt{25 - 20} = \sqrt 5$. Closed interval method: $P(0) = 10$ (the vertical segment, counted twice), $P(5) = 20$ (the diameter, counted twice) and $P(2\sqrt 5) = 8\sqrt 5 + 2\sqrt 5 = 10\sqrt 5$. Exact comparison: $10\sqrt 5 > 20$ because $\sqrt 5 > 2$, since $5 > 4$. The maximum perimeter is $10\sqrt 5$, for the rectangle $4\sqrt 5$ by $\sqrt 5$. Here an endpoint value is NOT zero, and skipping the comparison would leave the answer unproved.

e) For the area, base $5\sqrt 2$ and height $\frac{5\sqrt 2}{2}$: the base is TWICE the height, so the rectangle is two squares side by side. For the perimeter, base $4\sqrt 5$ and height $\sqrt 5$: the base is FOUR times the height, a flatter rectangle, because the perimeter rewards width more than the area does. The same figure, two objective functions, two answers: the first line of every solution must say which quantity is optimized.

a) The outline is two vertical sides $h$, the bottom $2r$ and the half-circle $\pi r$: $2h + 2r + \pi r = 8$, so $h = \frac{8 - (2 + \pi)r}{2}$. The area is $A = 2rh + \frac{\pi r^2}{2}$, and substituting, $A(r) = r\left(8 - (2 + \pi)r\right) + \frac{\pi r^2}{2} = 8r - 2r^2 - \pi r^2 + \frac{\pi r^2}{2} = 8r - \frac{4 + \pi}{2}r^2$. Domain: $r \ge 0$ and $h \ge 0$, that is $(2 + \pi)r \le 8$, so $r \in \left[0, \frac{8}{2 + \pi}\right]$. At the right endpoint, $h = 0$: the window is a half-disk alone, a real window with a real area.