Question

The equal sides of an isosceles triangles are each 8 in. long, the base being variable. Show that the triangle of maximum area is the one which has a right angle. Take one of the base angles as the independent variable, \(\phi\).

Short answer

Answer: The maximum area is achieved when the angle between one of the equal sides and the base is 45 degrees, resulting in a right angle at the midpoint of the base.

Step-by-step solution

Express the base and height in terms of \(\phi\)

Let's label the triangle's vertices A, B, and C, where A and C are the points where the equal sides meet, and B is the midpoint of the base. We can then use trigonometry to express the base BC and height (the perpendicular from A to BC) in terms of \(\phi\). Using the sine function, we can write the height (AS) as: $$ AS = 8 \sin \phi $$ Similarly, using the cosine function, we can express half of the base (BS) as: $$ BS = 8 \cos \phi $$ The entire base (SC) would be twice this value, so: $$ SC = 16 \cos \phi $$

Find the area of the triangle in terms of \(\phi\)

Now that we have the base and height in terms of \(\phi\), we can find the area of the triangle. The area of a triangle is given by the formula: $$ Area = \frac{1}{2} \times base \times height $$ Substituting the expressions for SC and AS, we get: $$ Area(\phi) = \frac{1}{2} \times (16 \cos \phi) \times (8 \sin \phi) = 64 \sin \phi \cos \phi $$

Find the maximum area using calculus

To find the maximum area, we will differentiate the area function with respect to \(\phi\), and set the derivative equal to zero. The derivative of the area function with respect to \(\phi\) is: $$ \frac{d(Area)}{d\phi} = 64 (\cos^2 \phi - \sin^2 \phi) $$ Setting the derivative to zero, we get: $$ \cos^2 \phi - \sin^2 \phi = 0 $$

Show that the maximum area is achieved when \(\phi = 90^\circ\)

To show that the maximum area is achieved when \(\phi = 90^\circ\), we will find the value of \(\phi\) for which the above equation holds true. Using the identity \(\sin^2 \phi + \cos^2 \phi = 1\), we can rewrite the equation as: $$ 2 \cos^2 \phi = 1 $$ This gives us: $$ \cos^2 \phi = \frac{1}{2} $$ Which means: $$ \cos \phi = \frac{1}{\sqrt{2}} $$ This is true for \(\phi = 45^\circ\) or \(\phi = 135^\circ\). However, we will consider only the \(\phi = 45^\circ\) value, since an isosceles triangle can only have angles less than 180 degrees. Since \(\phi = 45^\circ\), our triangle is a right-angled triangle with a right angle at B. This means that the triangle with maximum area has a right angle at the base.

Key concepts

Maximum Area

In geometry, maximizing the area of a figure is a common problem. For a given isosceles triangle, this task often involves tweaking its dimensions so the area reaches its largest possible value. The area of a triangle can be calculated using the formula: \[Area = \frac{1}{2} \times base \times height\]
By changing the base and height within the confines of the isosceles properties, we aim to maximize the expression while respecting geometric boundaries. In the context of this exercise, if the equal sides are fixed at 8 inches, we can maximize the area by altering the base angle \(\phi\) . Using trigonometry, this exploration leads us to find that when the triangle incorporates a right angle, the area is maximized. This scenario provides the optimal condition for expanding the height efficiently without altering fixed side lengths.

Trigonometry

Trigonometry plays a crucial role in determining the dimensions of an isosceles triangle in terms of angles. In this problem, we use the base angle \(\phi\) to express other side lengths.

• The height \(AS\) can be expressed as \(8 \sin \phi\), using the sine function which relates the opposite side to the hypotenuse in a right triangle.
• Half of the base \(BS\) is \(8 \cos \phi\), utilizing the cosine function to express the adjacent side in terms of angle \(\phi\) and hypotenuse.

Thus, the full base SC, being double of BS, becomes \(16 \cos \phi\).
These trigonometric relations provide a way to reframe the base and height into expressions involving \(\phi\), allowing us to dive into calculus for investigating the maximum area.

Calculus

Calculus helps in determining where the maximum or minimum of a function occurs by using derivatives. In this triangular problem, the area function involving \(\phi\) needs optimization. The function for area\[Area(\phi) = 64 \sin \phi \cos \phi\] can be analyzed through its derivative:
\[\frac{d(Area)}{d\phi} = 64 (\cos^2 \phi - \sin^2 \phi)\] Set this derivative to zero to find the critical points. Solving:\[\cos^2 \phi - \sin^2 \phi = 0\]Using trigonometric identities, notably \(\sin^2 \phi + \cos^2 \phi = 1\), simplifies finding \(\phi\) values that lead to maximum area configuration. Hence, calculus confirms the optimal area condition arises when \(\phi = 45^\circ\), forming a 90-degree angle at the triangle's base.
Thus, calculus verifies how geometry and trigonometry interplay to maximize area effectively.