Example 25: Trapezium with three 10 cm sides: area at most $75\sqrt3$

Basics you need

Solution

1. Set up the figure: QB equals AP

   Draw DP and CQ perpendicular to AB, and let AP = $x$ cm. Since DC is parallel to AB, DP = CQ (both are the distance between the parallel lines), and DPQC is a rectangle, so PQ = DC = 10 cm. The right triangles APD and BQC have equal hypotenuses (AD = BC = 10) and equal legs (DP = CQ), so they are congruent and

   $$\text{QB} = \text{AP} = x, \qquad \text{AB} = x + 10 + x = 2x + 10.$$
   We need the base AB in terms of $x$⁠. The book says the triangles are similar; because their hypotenuses are equal, they are in fact congruent, which is exactly what gives QB = $x$⁠. The picture follows the book in taking AB as the longer parallel side, with the slanting sides leaning outward; Step 9 shows the other possible shapes do worse.
   RHS congruence (Basic 2). Opposite sides of a rectangle are equal.

2. The height and the allowed range

   In the right triangle APD, by Pythagoras,

   $$\text{DP} = \sqrt{\text{AD}^2 - \text{AP}^2} = \sqrt{100 - x^2}.$$

   For a genuine trapezium of this shape we need $x > 0$ (otherwise the slanting sides are vertical) and $x < 10$ (otherwise the height is 0). So $0 < x < 10$⁠.

   The height is the other quantity in the area formula. The range tells us which values of $x$ are allowed, and in particular rules out $x = -10$ later.
   Pythagoras' theorem. Basics: area and volume formulas

3. The area as a function of $x$

   $$\begin{aligned} A(x) &= \tfrac12\,(\text{AB} + \text{DC})\cdot \text{DP} \\ &= \tfrac12\,(2x + 10 + 10)\sqrt{100 - x^2} \\ &= (x + 10)\sqrt{100 - x^2}. \end{aligned}$$
   This is the quantity to maximise, now in one variable. Halving $2x + 20$ gives $x + 10$⁠.
   Area of a trapezium (Basic 1). Basics: building the function to optimise

4. Differentiate: product rule and chain rule

   Take $u = x + 10$ and $v = \sqrt{100 - x^2}$⁠, so $u' = 1$ and, by the chain rule, $v' = \frac{-x}{\sqrt{100 - x^2}}$⁠. Then

   $$\begin{aligned} A'(x) &= \sqrt{100 - x^2} + (x+10)\frac{-x}{\sqrt{100 - x^2}} \\ &= \frac{(100 - x^2) - x^2 - 10x}{\sqrt{100 - x^2}} \\ &= \frac{-2x^2 - 10x + 100}{\sqrt{100 - x^2}} \\ &= \frac{-2(x + 10)(x - 5)}{\sqrt{100 - x^2}}. \end{aligned}$$
   Putting both terms over the common denominator $\sqrt{100 - x^2}$ makes the numerator a quadratic, which we can factorise to see where $A' = 0$ and what sign $A'$ has.
   Product rule; chain rule with inner function $100 - x^2$ (Basic 3). $-2x^2 - 10x + 100 = -2(x^2 + 5x - 50)$ (Basic 5).

5. The critical point

   The denominator is positive on $0 < x < 10$⁠, so

   $$A'(x) = 0 \iff (x + 10)(x - 5) = 0 \iff x = 5 \text{ or } x = -10.$$

   $x = -10$ is outside the range (a length cannot be negative), so the only critical point is $x = 5$⁠.

   The book solves $2x^2 + 10x - 100 = 0$⁠, the same equation multiplied by $-1$⁠. A fraction is zero exactly when its numerator is zero and its denominator is not.
   A product is zero when one factor is zero. Basics: factorising and the discriminant

6. Second derivative test at $x = 5$

   Write $A' = \frac{N}{\sqrt{100 - x^2}}$ with $N = -2x^2 - 10x + 100$⁠, $N' = -4x - 10$⁠. By the quotient rule,

   $$\begin{aligned} A''(x) &= \frac{N'(100 - x^2) + xN}{(100 - x^2)^{3/2}} \\ &= \frac{2x^3 - 300x - 1000}{(100 - x^2)^{3/2}}. \end{aligned}$$

   At $x = 5$⁠, $(100 - 25)^{3/2} = 75\sqrt{75}$⁠, so

   $$A''(5) = \frac{250 - 1500 - 1000}{75\sqrt{75}} = \frac{-30}{\sqrt{75}} < 0.$$

   By the second derivative test, $x = 5$ is a point of local maxima.

   The book's "on simplification" hides a little algebra: multiplying the quotient-rule numerator and denominator by $\sqrt{100 - x^2}$ gives $N'(100 - x^2) + xN$ over $(100 - x^2)^{3/2}$⁠, and expanding $N'(100-x^2) + xN$ gives $2x^3 - 300x - 1000$⁠. ($-\frac{30}{\sqrt{75}} = -2\sqrt3$⁠.)
   Quotient rule and $\frac{d}{dx}\sqrt{100 - x^2} = \frac{-x}{\sqrt{100 - x^2}}$⁠. Theorem 4(i), Section 6.4. Lesson: first and second derivative tests

7. It is the largest area on the whole range

   In $A'(x) = \frac{-2(x+10)(x-5)}{\sqrt{100-x^2}}$⁠, the factors $-2$⁠, $x + 10$ and the denominator have fixed signs on $0 < x < 10$⁠; only $x - 5$ changes sign. So $A' > 0$ on $0 < x < 5$ and $A' < 0$ on $5 < x < 10$⁠: $A$ increases up to $x = 5$ and decreases after it. Hence $A(x) < A(5)$ for every other allowed $x$⁠. Near the ends, $A$ approaches $A(0) = 100$ (a 10 cm square) and $A(10) = 0$ (a flat figure), both smaller.

   The second derivative test alone only compares nearby shapes. The sign of $A'$ on the whole range shows that no other lean of the sides gives more area.
   Theorem 1, Section 6.3: $A' > 0$ gives increasing, $A' < 0$ gives decreasing. Basics: sign of a product

8. The maximum area

   $$\begin{aligned} A(5) &= (5 + 10)\sqrt{100 - 25} = 15\sqrt{75} \\ &= 15 \cdot 5\sqrt3 = 75\sqrt3 \text{ cm}^2. \end{aligned}$$

   At this shape the base is AB = 20 cm, the height is $5\sqrt3$ cm, and $\cos A = \frac{5}{10} = \frac12$⁠, so the base angles are $60^\circ$⁠: the best trapezium is half of a regular hexagon of side 10 cm.

   The question asks for the area, so we evaluate $A$ at $x = 5$⁠. The hexagon remark is a neat way to remember the answer.
   $\sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt3$⁠.

9. The other possible shapes do worse

   The book's figure assumes the base is the longer parallel side. The other cases give less area:

   • Base shorter than DC (sides leaning inward by $x$⁠, $0 < x < 5$⁠): area $(10 - x)\sqrt{100 - x^2} < 10 \cdot 10 = 100$⁠.
   • Both slanting sides leaning the same way: AB = 10, a parallelogram with area $10 \times \text{height} \le 100$⁠.

   Both are at most 100, below $75\sqrt3 \approx 129.9$⁠.

   "Three sides equal to 10" allows these shapes too, so a complete answer should rule them out. Each is a product of two factors, each at most 10.

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