Optimisation — Word Problems & AM-GM

1. Express the target $Q$ and the constraint. 2. Eliminate a variable so $Q=Q(t)$ in one variable. 3. Solve $Q'(t)=0$ for critical $t$. 4. Verify with $Q''$ or the first-derivative test (and check the domain's endpoints). Classic set-ups: a cylinder/box of fixed volume with least surface, the shortest distance from a point to a curve, $\sum (x-a_j)^2$ minimised at the mean. Shortest distance from a point to a curve — parametrise the curve. On $y^2=4ax$ use $(at^2,2at)$; on $x^2=4ay$ use $(2at,at^2)$; on a circle $(r\cos\theta,r\sin\theta)$. Minimise the SQUARED distance $D(t)=(x(t)-x_0)^2+(y(t)-y_0)^2$ (same minimiser, no square root), and check the answer: at the nearest point the segment is NORMAL to the curve. For $y^2=4x$ and $(a,0)$ with $a>2$: $t^2=a-2$, distance $2\sqrt{a-1}$; for $a\le2$ the vertex is nearest.

Model the area/volume, use the geometric constraint (perimeter, radius, given side), reduce to one variable (often an angle $\theta$), and optimise. Standard results to quote:

• Max-area triangle inscribed in a circle of radius $R$ is equilateral, area $\tfrac{3\sqrt3}{4}R^2$; max-area rectangle inscribed in a circle of radius $r$ is a square, area $2r^2$ (sides $r\sqrt2$).
• Fixed-perimeter rectangle ⇒ square; fixed-perimeter triangle of greatest area ⇒ equilateral (side $\tfrac P3$, altitude $\tfrac{\sqrt3}{2}\cdot\tfrac P3$, area $\tfrac{\sqrt3}{36}P^2$).
• Sector of radius $r$, angle $\theta$ (radians): arc $l=r\theta$, perimeter $2r+r\theta$, area $\tfrac12r^2\theta=\tfrac12rl$. For fixed perimeter $P$ the area $\tfrac12r(P-2r)$ is a quadratic in $r$: max at $r=\tfrac P4$, i.e. $\theta=2$ radians, area $\tfrac{P^2}{16}$.
• Mensuration the set-ups borrow: trapezium area $\tfrac12(a+b)h$ (the max-area trapezium inscribed in a semicircle of radius $r$ has its top side $r$ and area $\tfrac{3\sqrt3}{4}r^2$); right triangle area $\tfrac12\times$legs; cylinder $V=\pi r^2h$, $S=2\pi rh+2\pi r^2$; cone $V=\tfrac13\pi r^2h$; sphere $V=\tfrac43\pi r^3$.

Collapses to memorise, each followed by 'max = coefficient × 1':

• Harmonic form: $a\sin x+b\cos x=\sqrt{a^2+b^2}\,\sin(x+\alpha)$, so $a\sin x+b\cos x+c$ has maximum $c+\sqrt{a^2+b^2}$ and minimum $c-\sqrt{a^2+b^2}$. $\sin\theta+\cos\theta=\sqrt2\sin(\theta+\tfrac\pi4)$, maximum $\sqrt2$ attained where $\theta+\tfrac\pi4=\tfrac\pi2$.
• Double angle: $\sin x\cos x=\tfrac12\sin2x$ (max $\tfrac12$); $\sin2x\cos2x=\tfrac12\sin4x$ (max $\tfrac12$); $2\cos^2x-1=\cos2x$; $\sin^2x=\tfrac{1-\cos2x}{2}$ so $4\sin^2x+1$ has max 5 and min 1.
• Reciprocal pair: $\tan x+\cot x=\dfrac{1}{\sin x\cos x}=\dfrac{2}{\sin2x}\ge2$ on $(0,\tfrac\pi2)$, so $\dfrac{1}{\tan x+\cot x}=\tfrac12\sin2x$ has max $\tfrac12$.
• Triple angle: $3\sin x-4\sin^3x=\sin3x$, $4\cos^3x-3\cos x=\cos3x$ — range $[-1,1]$.
• Sum of squares of sine and cosine of the SAME angle is 1, so $a\sin^2x+b\cos^2x$ lies between $\min(a,b)$ and $\max(a,b)$.

For a quadratic in $\sin x$ or $\cos x$ that does not collapse, put $t=\sin x\in[-1,1]$ and use the vertex or the endpoints — still no derivative.

For positive terms, AM $\ge$ GM: $\dfrac{u+v}{2}\ge\sqrt{uv}$, equality at $u=v$. So a sum with fixed product is minimised, and a product with fixed sum is maximised, when the terms are equal. Examples: $a^2x+b^2y$ with $xy=c^2$ has min $2abc$; $x+y=k\Rightarrow xy$ is max at $x=y=k/2$.

• The squared face of AM–GM: $ab\le\dfrac{a^2+b^2}{2}$ (apply it to $a^2,b^2$), equality at $a=b$. So under $a^2+b^2=64$ the product $ab$ is at most 32, and a rectangle inscribed in a circle of radius $r$ ($a^2+b^2=4r^2$) has area at most $2r^2$ — a square.
• The other calculus-free move — the quadratic vertex: $ax^2+bx+c$ has its extreme at $x=-\dfrac{b}{2a}$, value $c-\dfrac{b^2}{4a}=-\dfrac{D}{4a}$ (a minimum if $a>0$, a maximum if $a<0$). Anything that reduces to a quadratic in one variable — including $\sin^2x$ or $\cos x$ as the variable — is finished this way, provided the vertex lies in the variable's allowed range (else use the endpoints).