Students will learn how to find extrema graphically.

Conclude by giving your students these challenges:

- Moving Squares by NRICH
- N Is a Number by NRICH
- 2002 AMC 8, Problem 14
- 2000 AMC 8, Problem 4

The Deca Tree by NRICH: Here's the solution: A decatree has 10 trunks per tree, 10 branches per trunk, 10 twigs per branch, and 10 leaves per twig. Thus, a decatree has \(10^4\) leaves. Removing one trunk removes \(1/10\) of the leaves, which is \(10^4/10 = 10^3\) leaves. Removing one branch removes \(1/100\) of the leaves, which is \(10^4/10^2 = 10^2\) leaves. Continuing in this way, we find the remaining number of leaves:

$$\begin{align} & 10^4 - 10^3 - 10^2 - 10^1 - 10^0 \\ =\ & 10000 - 1000 - 100 - 10 - 1 \\ =\ & 10000 - 1111 \\ =\ & 8999 \end{align}$$Calculations such \(10^4/10\) can be carried out by students without knowing the quotient of powers rule for exponents. Specifically, students can rewrite \(10^4/10\) as \(10{,}000/10,\) then reduce the fraction, or find the quotient, whichever method the student prefers.

Find the area of this triangle:

Our goal is to determine the base and height of the triangle. Draw a segment connecting B to AC, which is perpendicular to AC. Call the point of intersection E, and label BE with h, the height of the triangle. The definition of \(\sin\) tells us that \(\sin 2x = h / 18\) and \(\sin x = h / 30.\) Putting these together, we find

$$h = 18\sin 2x = 30\sin x$$Using the double angle formula for sine, we get

$$18 \cdot 2\sin x\cos x = 30\sin x$$Solving for \(\cos x,\) we get \(\cos x = 5/6.\) We now know \(\cos x,\) but our \(h\) is in terms of \(\sin x.\) From the Pythagorean identity \(\sin^2 x + \cos^2 x = 1,\) it's easy to write \(\sin\) in terms of \(\cos,\) like so:

$$\sin x = \pm\sqrt{1 - \cos^2 x}$$But since \(h = 30\sin x\) and \(h\) is a length, we know \(\sin x\) can't be negative. After some arithmetic, we find \(h = 5\sqrt{11}.\) Now we know the height of the triangle, but we still have to find the length of its base. Remember, we know that \(\cos x = 5/6.\) But using the definition of \(\cos,\) we also know that \(\cos x = EC/30,\) so \(5/6 = EC/30,\) and thus \(EC = 25.\) If we could just figure out AE, we would have the length of our base. Notice that \(x\) is less than \(2x.\) Thus, we can construct segment BD, such that \(\triangle BDA\) is isosceles. Since \(\triangle BDA\) is isosceles, \(BD = BA = 18.\) Next, notice that \(\angle BDA,\) which has measure \(2x,\) is an exterior angle of \(\triangle BDC.\) Thus \(\angle CBD = x,\) and so \(\triangle BDC\) is also isosceles. Because \(\triangle BDC\) is isosceles, we know \(DC = DB = 18.\) We know \(EC = ED + DC,\) but we also know \(EC = 25,\) and \(DC = 18.\) Thus, \(ED\) must be \(7.\) From this, it's obvious that \(AE\) is also \(7,\) and so our triangle has a base of length \(7 + 7 + 18 = 32.\) Finally, we plug our base and height into our area formula to get \((1/2)32 \cdot 5\sqrt{11} = 80\sqrt{11}.\)