Question

Closing Off the First Quadrant You are planning to close off a corner of the first quadrant with a line segment 20 units long running from \((a, 0)\) to \((0, b) .\) Show that the area of the triangle enclosed by the segment is largest when \(a=b\) .

Short answer

The triangle's area is maximum when \(a = b = 10\sqrt{2}\).

Step-by-step solution

Establish the relationship between 'a' and 'b'

Since the line segment forms a right triangle with the x and y axis, we can use Pythagoras' Theorem to establish a relationship between 'a' and 'b'. This gives us the formula \(a^2 + b^2 = 20^2 \). We can solve this equation for 'a' or 'b'.

Solve for 'b'

Let's solve for 'b' in terms of 'a'. Rearranging the equation we get \(b = \sqrt{400 - a^2}\).

Express the area of triangle in terms of 'a'

The area of the triangle is given by \(Area = 0.5ab = 0.5a*\sqrt{400 - a^2}\). This is the equation that we will maximize.

Find the derivative of the Area and set it to zero

To find the maximum point of the function we need to find the derivative of the function with respect to 'a', set it equal to zero and solve for 'a'. The derivative of \(Area = 0.5a*\sqrt{400 - a^2}\) is \(Area' = 0.5*\sqrt{400 - a^2} - 0.5*a*\frac{a}{\sqrt{400 - a^2}}\). Setting this equal to zero and solving for 'a' will give us the critical points.

Determine the maximum area

After calculating the derivative and setting it equal to zero, we find that \(a = 10\sqrt{2}\). Substituting \(a = 10\sqrt{2}\) to equation \(b = \sqrt{400 - a^2}\) we find that \(b = 10\sqrt{2}\). Hence, the area of the triangle is maximized when \(a = b = 10\sqrt{2}\).