Corrected exercises: Applied optimization (MATH 203)

a) The front and the back together cost $18x + 6x = 24x$ dollars, the two sides $2 \times 6y = 12y$ dollars. The constraint is $24x + 12y = 1440$; dividing by $12$ first keeps the numbers small: $2x + y = 120$, so $y = 120 - 2x$. The area is $A = xy = x(120 - 2x) = 120x - 2x^2$. Domain: $x \ge 0$ and $y = 120 - 2x \ge 0$, so $x \in [0, 60]$. Both endpoints are flat gardens of area $0$; including them makes the interval closed, and $A$, a polynomial, is continuous on it: the Closed Interval Method applies.

b) $A'(x) = 120 - 4x$, zero for $x = 30$, inside $[0, 60]$. Values: $A(0) = 0$, $A(60) = 0$, $A(30) = 30 \times 60 = 1800$. The largest garden is $30$ m along the street and $y = 120 - 60 = 60$ m deep, with an area of $1800$ m$^2$. The answer is the pair of dimensions AND the area: $x = 30$ alone answers nothing.

c) Now the constraint is the area, $xy = 1800$, so $y = \frac{1800}{x}$, and the cost is $C(x) = 24x + 12 \cdot \frac{1800}{x} = 24x + 21600x^{-1}$, on the domain $(0, \infty)$: any positive front works, and there is no endpoint. Power rule on the negative exponent: $C'(x) = 24 - 21600x^{-2} = 24 - \frac{21600}{x^2}$. Over the common denominator $x^2$: $C'(x) = \frac{24x^2 - 21600}{x^2} = \frac{24(x^2 - 900)}{x^2} = \frac{24(x - 30)(x + 30)}{x^2}$. This ONE factored fraction does all the work. On $(0, \infty)$, $x^2 > 0$ and $x + 30 > 0$, so the sign of $C'$ is the sign of $x - 30$: negative on $(0, 30)$, positive on $(30, \infty)$. $C$ decreases then increases on the WHOLE domain, so $C(30)$ is the absolute minimum (the root $x = -30$ is not in the domain). Then $y = 60$ m and $C(30) = 720 + 720 = 1440$ dollars.

d) The error is in the derivative of $\frac{21600}{x}$. Rewritten as $21600x^{-1}$, the power rule gives $21600 \times (-1)x^{-2} = -\frac{21600}{x^2}$: the exponent $-1$ comes down as a factor and carries its MINUS sign. The classmate lost it. Correctly, $C'(10) = 24 - \frac{21600}{100} = 24 - 216 = -192 < 0$. The conclusion was absurd before any derivative: as $x \to 0^+$, $\frac{21600}{x} \to \infty$, a garden $1$ cm wide and $18$ km deep needs an enormous amount of mesh. A sign check at one convenient point, here $x = 10$, catches the lost minus in five seconds.

e) Both problems give the same garden, $30$ m by $60$ m, with area $1800$ m$^2$ and cost $1440$ dollars. It is not a coincidence: if a garden of area $1800$ could be fenced for less than $1440$ dollars, the leftover money would enlarge it beyond $1800$ m$^2$, contradicting b). At the optimum, the front and back cost $24 \times 30 = 720$ dollars and the two sides $12 \times 60 = 720$ dollars: half of the budget each. In general, $A = x \cdot \frac{B - 24x}{12}$ has $A' = \frac{B - 48x}{12} = 0$ for $24x = \frac{B}{2}$, whatever the budget $B$.

a) The base has area $x^2$ and each of the four flaps $xh$: the constraint is $x^2 + 4xh = 2700$, so $h = \frac{2700 - x^2}{4x}$. Then $V = x^2h = x^2 \cdot \frac{2700 - x^2}{4x}$. Cancel ONE factor $x$ between $x^2$ and $4x$: $V = \frac{x(2700 - x^2)}{4} = \frac{2700x - x^3}{4} = 675x - \frac{x^3}{4}$. Domain: $x > 0$ and $h \ge 0$, that is $x^2 \le 2700$, $x \le \sqrt{2700} = 30\sqrt 3 \approx 51.96$. The simplified formula is a polynomial, so it also makes sense at $x = 0$ with value $0$ (no box); including that endpoint gives the closed interval $[0, 30\sqrt 3]$. For $x = 10$: $h = \frac{2700 - 100}{40} = 65$ cm, a tall narrow box.

b) $V'(x) = 675 - \frac{3x^2}{4} = \frac{2700 - 3x^2}{4} = \frac{3(900 - x^2)}{4} = \frac{3(30 - x)(30 + x)}{4}$. In the domain the only zero is $x = 30$ ($-30$ is rejected). Closed Interval Method: $V(0) = 0$; $V(30\sqrt 3) = 0$ since then $h = 0$; $V(30) = 675 \times 30 - \frac{27000}{4} = 20250 - 6750 = 13\,500$. The largest box has a base of $30$ cm by $30$ cm and a height $h = \frac{2700 - 900}{120} = 15$ cm, for $13\,500$ cm$^3$. Check the constraint: $900 + 4 \times 30 \times 15 = 900 + 1800 = 2700$.

c) With $V = 675x - x^2$, $V' = 675 - 2x = 0$ gives $x = 337.5$ cm, far beyond $30\sqrt 3 \approx 52$: that box would need a base of more than $11$ m$^2$ from $0.27$ m$^2$ of cardboard. A candidate outside the domain is often the first sign of an algebra slip. The slip: the denominator $4x$ divides the WHOLE numerator $2700 - x^2$, so $x^2 \cdot \frac{2700 - x^2}{4x} = \frac{2700x}{4} - \frac{x^3}{4}$. The classmate divided the first term by $4x$ and forgot to divide the second. Splitting a fraction is legal only term by term, and every term keeps the denominator.

d) Now there are two squares: $2x^2 + 4xh = 2700$, so $h = \frac{2700 - 2x^2}{4x} = \frac{1350 - x^2}{2x}$ and $V = x^2h = \frac{x(1350 - x^2)}{2} = 675x - \frac{x^3}{2}$ on $[0, \sqrt{1350}]$. $V'(x) = 675 - \frac{3x^2}{2} = \frac{3(450 - x^2)}{2}$, zero for $x = \sqrt{450} = 15\sqrt 2 \approx 21.21$ cm. $V$ vanishes at both endpoints and is positive in between, so this is the absolute maximum. Then $h = \frac{1350 - 450}{30\sqrt 2} = \frac{900}{30\sqrt 2} = \frac{30}{\sqrt 2} = 15\sqrt 2$: the height equals the side, the optimal closed box is a CUBE. $V = (15\sqrt 2)^3 = 3375 \times 2\sqrt 2 = 6750\sqrt 2 \approx 9546$ cm$^3$ (calculator, nearest cm$^3$). The lid costs material, so the volume drops from $13\,500$ to about $9546$ cm$^3$.

e) $x^2 + 4xh = M$ gives $h = \frac{M - x^2}{4x}$ and $V = \frac{Mx - x^3}{4}$. $V'(x) = \frac{M - 3x^2}{4}$, positive then negative around $x^2 = \frac{M}{3}$, with $V = 0$ at both endpoints: the absolute maximum. There, $M - x^2 = 3x^2 - x^2 = 2x^2$, so $h = \frac{2x^2}{4x} = \frac{x}{2}$. Replacing $M$ by $3x^2$ instead of computing decimals is the algebraic gesture that makes the shape appear. With $M = 2700$: $x = 30$, $h = 15$, as in b).

a) The cup has one disk of area $\pi r^2$ and a side of area $2\pi rh$: $A = \pi r^2 + 2\pi rh$. The constraint is $\pi r^2 h = 1000$, so $h = \frac{1000}{\pi r^2}$ (solve for $h$, never for $r$). Substituting, $2\pi r \cdot \frac{1000}{\pi r^2} = \frac{2000}{r}$, so $A(r) = \pi r^2 + \frac{2000}{r}$. Every $r > 0$ gives a cup, so the domain is $(0, \infty)$, open at $0$ and unbounded.

b) Rewrite $\frac{2000}{r} = 2000r^{-1}$: $A'(r) = 2\pi r - 2000r^{-2} = \frac{2\pi r^3 - 2000}{r^2} = \frac{2(\pi r^3 - 1000)}{r^2}$. On $(0, \infty)$, $r^2 > 0$, and $\pi r^3 - 1000$ is increasing, negative for small $r$ and positive for large $r$, zero once, at $r^3 = \frac{1000}{\pi}$, $r = \frac{10}{\sqrt[3]{\pi}} \approx 6.83$ cm. So $A$ decreases then increases on the WHOLE domain: the absolute minimum is at $r \approx 6.83$ cm. There $\pi r^3 = 1000$, so $\frac{2000}{r} = \frac{2\pi r^3}{r} = 2\pi r^2$ and $A = \pi r^2 + 2\pi r^2 = 3\pi r^2 \approx 439.38$ cm$^2$. Keep the exact $r$ in the calculator for this last step: rounding $r$ to $6.83$ first and then computing $3\pi r^2$ gives $439.66$, the wrong hundredth.

c) $h = \frac{1000}{\pi r^2}$. At the optimum, $1000 = \pi r^3$: substitute it in the numerator, $h = \frac{\pi r^3}{\pi r^2} = r$. The open cup is exactly as tall as its radius, half as tall as it is wide. Using the equation $A'(r) = 0$ to replace a number by an expression in $r$ is the algebraic gesture that turns a decimal answer into a shape.

d) Each disk now costs a square of area $(2r)^2 = 4r^2$, so $A = 2 \times 4r^2 + 2\pi rh = 8r^2 + \frac{2000}{r}$ on $(0, \infty)$. $A'(r) = 16r - \frac{2000}{r^2} = \frac{16r^3 - 2000}{r^2} = \frac{16(r^3 - 125)}{r^2}$, negative on $(0, 5)$ and positive on $(5, \infty)$: the absolute minimum is at $r = 5$ cm. Metal used: $A(5) = 200 + 400 = 600$ cm$^2$. Then $h = \frac{1000}{25\pi} = \frac{40}{\pi} \approx 12.73$ cm and $\frac{h}{r} = \frac{8}{\pi} \approx 2.55$. Paying for the corners makes the ends expensive, so the optimal can is taller and narrower than a can with $h = 2r$.

e) To clear the negative exponent, multiply both sides of $2\pi r = 2000r^{-2}$ by $r^2$, since $r^{-2} \cdot r^2 = 1$: $2\pi r^3 = 2000$, $r^3 = \frac{1000}{\pi}$. Multiplying by $r^{-2}$ instead makes the exponent MORE negative on the other side, and the result $r = \frac{\pi}{1000} \approx 0.003$ cm, a cup three hundredths of a millimetre wide, should have stopped the classmate. Better still: write $A'$ as one fraction first, as in b), and the equation becomes numerator $= 0$.

a) $D = x^2 + y^2$ with $y = 4 - x^2$: $D(x) = x^2 + (4 - x^2)^2$. Expand the square as a binomial, $(4 - x^2)^2 = 16 - 8x^2 + x^4$, never as $16 - x^4$ or $16 + x^4$: $D(x) = x^4 - 7x^2 + 16$. The domain is $\mathbb{R}$, since every $x$ gives a point of the parabola. $D(1) = 1 - 7 + 16 = 10$, which checks against the figure: $P(1, 3)$ is at distance $\sqrt{10}$ from $O$.

b) $D'(x) = 4x^3 - 14x$. Factor the common factor $2x$: $D'(x) = 2x(2x^2 - 7)$. A product is zero when one factor is zero: $x = 0$, or $x^2 = \frac{7}{2}$, $x = \pm\sqrt{\frac{7}{2}} = \pm\frac{\sqrt{14}}{2} \approx \pm 1.8708$. Three critical numbers. Dividing both sides of $4x^3 - 14x = 0$ by $x$ is only legal when $x \ne 0$, and it silently deletes the solution $x = 0$. The rule: move everything to one side and FACTOR; never divide an equation by an expression that can be zero.

c) Signs of the factors: $2x$ changes sign at $0$, and $2x^2 - 7$ is negative between $-\sqrt{\frac{7}{2}}$ and $\sqrt{\frac{7}{2}}$, positive outside. So $D' < 0$ on $\left(-\infty, -\sqrt{\frac{7}{2}}\right)$, $D' > 0$ on $\left(-\sqrt{\frac{7}{2}}, 0\right)$, $D' < 0$ on $\left(0, \sqrt{\frac{7}{2}}\right)$ and $D' > 0$ on $\left(\sqrt{\frac{7}{2}}, \infty\right)$. $D$ decreases, increases, decreases, increases, and $D \to \infty$ as $x \to \pm\infty$: the absolute minimum is the smaller of the two local minima, and they are equal by symmetry, $D\left(\pm\sqrt{\frac{7}{2}}\right) = \frac{49}{4} - \frac{49}{2} + 16 = \frac{15}{4}$. The closest points are $\left(\pm\frac{\sqrt{14}}{2}, \frac{1}{2}\right)$, since $y = 4 - \frac{7}{2} = \frac{1}{2}$, and the least distance is $\sqrt{\frac{15}{4}} = \frac{\sqrt{15}}{2} \approx 1.9365$. At $x = 0$, $D'$ goes from positive to negative: by the First Derivative Test it is a local MAXIMUM of $D$, the vertex at distance $4$. A critical number is a candidate, not an answer.

d) On the closed interval $[-2, 2]$, the continuous $D$ reaches a maximum, and the Closed Interval Method compares ALL the critical numbers with the endpoints: $D(-2) = 16 - 28 + 16 = 4$, $D\left(\pm\sqrt{\frac{7}{2}}\right) = \frac{15}{4}$, $D(0) = 16$, $D(2) = 4$. The farthest point of the arch is the vertex $(0, 4)$, at distance $\sqrt{16} = 4$. The classmate of b), who lost $x = 0$, compares only $\frac{15}{4}$ and $4$ and answers the endpoints $(\pm 2, 0)$, at distance $2$: half the true answer, lost in one illegal division.

e) $D_c(x) = x^2 + (4 - c - x^2)^2$. Chain rule on the square, inner function $4 - c - x^2$: $D_c'(x) = 2x + 2(4 - c - x^2)(-2x) = 2x\left[1 - 2(4 - c - x^2)\right] = 2x(2x^2 + 2c - 7)$. If $c \ge \frac{7}{2}$, then $2x^2 + 2c - 7 \ge 2x^2 > 0$ for $x \ne 0$, so $D_c'$ has the sign of $x$: $D_c$ decreases then increases, and the vertex is the closest point. If $c < \frac{7}{2}$, the second factor vanishes at $x = \pm\sqrt{\frac{7 - 2c}{2}}$ and, as in c), $x = 0$ becomes a local maximum. So the vertex is closest exactly when $c \ge \frac{7}{2}$; $c = 0$ gives back b).

a) Above the rectangle sits a small right triangle with legs $x$ (horizontal) and $9 - y$ (vertical), similar to the big one, whose legs are $12$ and $9$. Corresponding sides: $\frac{9 - y}{x} = \frac{9}{12} = \frac{3}{4}$, so $9 - y = \frac{3x}{4}$ and $y = 9 - \frac{3x}{4}$, which is also the equation of the hypotenuse. Then $A(x) = xy = 9x - \frac{3x^2}{4}$, on $[0, 12]$ (at both ends the rectangle is flat). The ratio must pair a vertical side with a vertical side: writing $\frac{9 - y}{x} = \frac{12}{9}$ gives $y = 9 - \frac{4x}{3}$, which reaches $0$ at $x = 6.75$ instead of $12$, a check that fails at once.

b) $A'(x) = 9 - \frac{3x}{2} = \frac{3(6 - x)}{2}$, zero at $x = 6$. Closed Interval Method: $A(0) = A(12) = 0$ and $A(6) = 54 - 27 = 27$. The largest corner rectangle is $6$ by $\frac{9}{2}$, area $27$: exactly half of the triangle, whose area is $\frac{12 \times 9}{2} = 54$. Its corner is the midpoint of the hypotenuse.

c) Area of the triangle two ways: $\frac{1}{2} \times 15 \times H = 54$, so $H = \frac{108}{15} = \frac{36}{5} = 7.2$. The side of the rectangle opposite the hypotenuse is parallel to it, at distance $7.2 - t$ from the right angle: it cuts off a small triangle similar to the big one in the ratio $\frac{7.2 - t}{7.2}$. Hence $w = 15 \cdot \frac{7.2 - t}{7.2} = 15 - \frac{15t}{7.2}$. Clear the decimal: $\frac{15}{7.2} = \frac{150}{72} = \frac{25}{12}$, so $w = 15 - \frac{25t}{12}$ and $A(t) = tw = 15t - \frac{25t^2}{12}$ on $[0, 7.2]$. $A'(t) = 15 - \frac{25t}{6} = 0$ for $t = \frac{90}{25} = 3.6$, with $A(0) = A(7.2) = 0$. So $t = 3.6$, $w = 15 - 7.5 = 7.5$ and $A = 3.6 \times 7.5 = 27$. The figure's rectangle has its far corners at $(0, 4.5)$ and $(6, 0)$: $\sqrt{36 + 20.25} = 7.5$, as computed.

d) Triangle of base $b$ and height $H$, rectangle of height $t$ standing on the base. Its top side cuts off a triangle similar to the whole in the ratio $\frac{H - t}{H}$, so $w = \frac{b(H - t)}{H}$ and $A(t) = \frac{b}{H}(Ht - t^2)$ on $[0, H]$. $A'(t) = \frac{b}{H}(H - 2t) = 0$ for $t = \frac{H}{2}$, with $A = 0$ at both ends: the maximum is $\frac{b}{H} \cdot \frac{H^2}{4} = \frac{bH}{4}$, half of the triangle's $\frac{bH}{2}$. In b), the base is the leg $12$ and $H = 9$: $t = 4.5$; in c), the base is the hypotenuse $15$ and $H = 7.2$: $t = 3.6$. Both give $\frac{54}{2} = 27$.