Exercises — Optimization — constrained, unconstrained, real-world problems

What we do: find where the slope is zero. Why: critical points are the only candidates for extrema. $f^{\prime }(x)=3x^2-3=3(x^2-1)=3(x-1)(x+1).$ Set $f^{\prime }(x)=0$: $x=1$ or $x=-1$. The slope is a polynomial, so it never "fails to exist" — those two are the complete list. Answer: critical points at $==x=-1==$ and $==x=1==$.

What we do: compute $f^{\prime \prime }$ and read its sign at each critical point. Why: the sign of $f^{\prime \prime }$ says cup (min) or cap (max), straight from the Taylor argument $f(c+h)-f(c)\approx \frac{1}{2}{f}^{\prime \prime }(c)h^2$. $f^{\prime \prime }(x)=6x.$

• At $x=-1$: $f^{\prime \prime }(-1)=-6<0$ → cap-shaped → local maximum.
• At $x=+1$: $f^{\prime \prime }(1)=+6>0$ → cup-shaped → local minimum. Answer: $x=-1$ is the local max, $x=1$ is the local min.

What we do: apply the Closed Interval Method — compare $f$ at every interior critical point and both endpoints. Why: on a closed interval the global extreme can hide at an endpoint even if the slope there is not zero.

• Critical points of $f^{\prime }(x)=3(x-1)(x+1)$ are $x=\pm 1$; only $x=1$ lies in $[0,2]$.
• Evaluate the candidates: $f(0)=0,f(1)=1-3=-2,f(2)=8-6=2.$ Answer: absolute maximum $==f(2)=2==$ at $x=2$ (an endpoint!); absolute minimum $==f(1)=-2==$ at $x=1$.

Draw & name (D, U): as in the figure, let $x$ = each of the two red sides perpendicular to the river, $y$ = the lavender side parallel to it. Area $A=xy$. Constrain to one variable (C): only three sides are fenced (the river side, dashed in the figure, is free), so $2x+y=40\Rightarrow y=40-2x$. $A(x)=x(40-2x)=40x-2x^2,0\le x\le 20.$ Why the domain? $x\ge 0$ is a real length; $x\le 20$ keeps $y=40-2x\ge 0$. Kill the derivative (K): $A^{\prime }(x)=40-4x=0\Rightarrow x=10.$ Discriminate (D): $A^{\prime \prime }(x)=-4<0$ → cap → maximum. ✓ Check endpoints (C): $A(0)=0$, $A(20)=0$, $A(10)=10\cdot 20=200$. Answer: $x=10$ m, $y=40-20=20$ m, giving maximum area $==200{\text{m}}^2==$.

What we do & why: we minimise the squared distance $S$ instead of the distance $D$. Since $\sqrt{\cdot }$ only ever increases, whatever $x$ makes $S$ smallest makes $D$ smallest too — and squaring kills the ugly square root. $S(x)=(x-0{)}^2+(y-3{)}^2=x^2+(x^2-3{)}^2.$ Expand: $S(x)=x^2+x^4-6x^2+9=x^4-5x^2+9$. Kill the derivative: $S^{\prime }(x)=4x^3-10x=2x(2x^2-5)=0.$ Roots: $x=0$ or $x^2=\frac{5}{2}\Rightarrow x=\pm \sqrt{5/2}$. Discriminate: $S^{\prime \prime }(x)=12x^2-10$.

• At $x=0$: $S^{\prime \prime }(0)=-10<0$ → a local max of squared distance (the butter point in the figure). Reject.
• At $x=\pm \sqrt{5/2}$: $S^{\prime \prime }>0$ → minima (the mint points). Global check on an unbounded domain: here $x$ ranges over all of $\mathbb{R}$, which has no endpoints — so we must rule out any escape to smaller values far away. As $|x|\to \infty$, the leading $x^4$ term dominates, so $S(x)\to +\infty$. The squared distance therefore grows without bound in both directions and cannot beat the interior minima. Since the two symmetric minima give equal, finite values and everything else is larger, they are the global minima. Compute the minimum distance-squared: $S(\pm \sqrt{5/2})=\frac{25}{4}-5\cdot \frac{5}{2}+9=\frac{25}{4}-\frac{50}{4}+\frac{36}{4}=\frac{11}{4}$. With $x^2=\frac{5}{2}$, $y=x^2=\frac{5}{2}$. Answer: two closest points $\left(\pm \sqrt{5/2},\frac{5}{2}\right)$, each at distance $==\sqrt{11/4}=\frac{\sqrt{11}}{2}\approx 1.658==$.

Name & model (U): surface area of a closed cylinder is $A=\underset{\text{two circular ends}}{\underbrace{2\pi r^2}}+\underset{\text{curved side}}{\underbrace{2\pi rh}}.$ Constrain to one variable (C): the volume constraint $\pi r^{2}h=1000$ gives $h=\frac{1000}{\pi r^2}$. Substitute: $A(r)=2\pi r^2+2\pi r\cdot \frac{1000}{\pi r^2}=2\pi r^2+\frac{2000}{r},r>0.$ Kill the derivative (K): $A^{\prime }(r)=4\pi r-\frac{2000}{r^2}=0\Rightarrow 4\pi r^3=2000\Rightarrow r^3=\frac{500}{\pi }.$ $r={\left(\frac{500}{\pi }\right)}^{1/3}\approx 5.419\text{cm}.$ Discriminate (D): $A^{\prime \prime }(r)=4\pi +\frac{4000}{r^3}>0$ for all $r>0$ → cup → minimum. ✓ (No endpoints to worry about since $r\to 0^{+}$ and $r\to \infty$ both send $A\to \infty$.) Height: $h=\frac{1000}{\pi r^2}$. Using $r^3=500/\pi$ we get $h=\frac{1000}{\pi r^2}=\frac{1000r}{\pi r^3}=\frac{1000r}{500}=2r\approx 10.839\text{cm}$. Answer: $r\approx ==5.419==\text{cm}$, $h=2r\approx ==10.839==\text{cm}$. Beautifully, the optimal can has height equal to its diameter ($h=2r$).

Route 1 — Lagrange multipliers. We want the largest $f$ while stuck on the plane $g(x,y,z)=x+y+z-12=0$. Geometrically the optimum is where $\nabla f$ points perpendicular to the constraint surface, i.e. parallel to $\nabla g$: $\nabla f=\lambda \nabla g.$ Here $\nabla f=(yz,xz,xy)$ and $\nabla g=(1,1,1)$, so $yz=\lambda ,xz=\lambda ,xy=\lambda .$ From $yz=xz$ (and $z>0$) we get $y=x$. From $xz=xy$ (and $x>0$) we get $z=y$. So $x=y=z$. The constraint forces $3x=12\Rightarrow x=y=z=4$, giving $f_{\max }=4\cdot 4\cdot 4===64==.$

Route 2 — AM–GM check. The Arithmetic Mean–Geometric Mean inequality says, for positive numbers, $\frac{x+y+z}{3}\ge \sqrt[3]{xyz},\text{equality iff }x=y=z.$ Here the left side is fixed: $\frac{12}{3}=4$. So $\sqrt[3]{xyz}\le 4\Rightarrow xyz\le 64$, with equality exactly when $x=y=z=4$. Same answer, no calculus. ✓

Why two routes agree: Lagrange found the critical point of a smooth function; AM–GM proved it is the global maximum. Together they show $64$ is not just a candidate — it is the true maximum.

• L1 — can you spot the flat spots and tell a hill from a dip?
• L2 — can you turn a story into a single-variable area (or cost) function and crank the recipe (watch the endpoints)?
• L3 — can you choose a smarter target ($D^2$) and refuse to trust the obvious point?
• L4 — can you use the constraint to collapse two variables into one?
• L5 — can you solve the same thing two ways and prove it's really the extreme?