Geometric optimisation with a trigonometric model

Trigonometric geometric optimisation

A geometric constraint can make the side lengths functions of an angle. Substitute those lengths into an area model, differentiate with respect to the angle and compare the relevant values

1. Relate lengths to the angle Use the given geometry to express each changing side length in terms of the angle before forming the area model \(\displaystyle L_1=8\sin u,\quad L_2=8\cos u\)
2. Use the area formula For a triangle, take half the product of the two side lengths and the sine of their included angle \(\displaystyle A_{\triangle}=\frac{1}{2}ab\sin u\)
3. Rewrite the trigonometric product Use a double-angle identity when the model contains the product of sine and cosine \(\displaystyle \sin u\cos u=\frac{1}{2}\sin(2u)\)
4. Test the optimum Find stationary angles from the derivative and compare their model values with any required endpoints \(\displaystyle A'(u)=0\quad\text{and compare }A(u)\)

Worked example

A shape consists of a fixed region and a right triangle with the dimensions below. Find the maximum total area

1. \(\displaystyle a=8\sin u,\quad b=8\cos u\)Resolve the two legs of the right triangle using the hypotenuse and angle u.
2. \(\displaystyle A(u)=3+\frac{1}{2}(8\sin u)(8\cos u)=3+32\sin u\cos u=3+16\sin(2u)\)Use the triangle area formula and the double-angle identity to form the total area
3. \(\displaystyle A'(u)=32\cos(2u)=0\quad\Rightarrow\quad 2u=\frac{\pi}{2}\quad\Rightarrow\quad u=\frac{\pi}{4}\)Set the derivative to zero to find the stationary angle in the interval
4. \(\displaystyle A\left(\frac{\pi}{4}\right)=3+16\sin\left(\frac{\pi}{2}\right)=19\text{ cm}^2\)The stationary value is 19 cm², greater than the endpoint value 3 cm², so it is the maximum

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