find the largest area of an isosceles triangle inscribed in a circle of radius 3

Answer:

Here is the answer and explanation all together

The largest area of an isosceles triangle that is inscribed in a circle of radius 3 units is 3.897 sq. units approx.

To find the maximum of a continuous and twice differentiable function f(x), we can firstly differentiate it with respect to x and equating it to 0 will give us critical points.

Putting those values of x in the second rate of function, if results in negative output, then at that point, there is maxima. If the output is positive then its minima and if its 0, then we will have to find the third derivative (if it exists) and so on.

For this case, we can construct a general circle and a variable isosceles triangle in it.

Let the radius of the circle be 'r' units.

For isosceles triangle inscribed in that circle, its vertex(intersection of two congruent sides) be fixed, and other two vertex are allowed to move equally on both the sides of that first vertex.

Those rest two vertices make a chord in that circle. Let the perpendicular distance from the center be 'd' units of that chord.

Then, by Pythagoras theorem for triangle OBD, we get:

[tex]|OB|^2 = |OD|^2 + |BD|^2\\|BD| = \sqrt{|OB|^2 - |OD|^2}\\|BD| = \sqrt{r^2 - d^2} \: \rm units[/tex]

Thus, as perpendicular from center to a chord bisects it, we get:

[tex]|BC| = |BD| + |DC| = |BD| + |BD| = 2\sqrt{r^2 - d^2}[/tex]

The distance from the first vertex (A) to the mid of the line segment BC is: r - d units.

Thus, area of the considered isosceles triangle is:

[tex]A = f(r,d) = \dfrac{1}{2} \times (r - d) \times 2\sqrt{r^2 - d^2} = (r-d)^{1.5}(r+d)^{0.5}[/tex]

At r = 3 units, we get:

[tex]f(3,d) = g(d) = (3-d)^{1.5}(3+d)^{0.5}[/tex]

Finding the value of 'd' for which the function g(d) becomes maximum will give us the maximum area.

Finding first and second derivative of g(d) with respect to 'd', we get:

[tex]g(d) = (3-d)^{1.5}(3+d)^{0.5}\\g'(d) = -1.5(3-d)^{0.5}(3+d)^{0.5} + 0.5(3+d)^{-0.5}(3-d)^{1.5}\\g''(d) = -1.5[-0.5(3-d)^{-0.5}(3+d)^{0.5}] + 0.5[-0.5(3+d)^{-1.5}(3-d)^{1.5}]\\[/tex]

Putting first rate = 0 to find the critical points, we get:

[tex]g'(d) =- 1.5(3-d)^{0.5}(3+d)^{0.5} + 0.5(3+d)^{-0.5}(3-d)^{1.5} = 0\\3(3-d)^{0.5}(3+d)^{0.5} = (3+d)^{-0.5}(3-d)^{1.5}\\3(3+d) = (3-d)\\9 + 3d = 3 - d\\d = 1.5[/tex]

At d = 1.5,  the second rate of g(d) evaluates to:

[tex]g''(1.5) = -1.5[-0.5(3-1.5)^{-0.5}(3+1.5)^{0.5}] + 0.5[-0.5(3+1.5)^{-1.5}(3-1.5)^{1.5}]\\g''(5) = 0.75(1.5)^{1.5}(4.5)^{0.5} -0.25(4.5)^{-1.5}(1.5)^{1.5}\\g''(5) =[/tex]

The area of the isosceles triangle is maximum, and is evaluated as:

[tex]A_{max} = f(3,1.5) = g(1.5) = (3-1.5)^{1.5}(3+1.5)^{0.5} \approx 3.897 \: \rm unit^2[/tex]

Thus, the largest area of an isosceles triangle that is inscribed in a circle of radius 3 units is 3.897 sq. units approx.

Learn more about maximum of a function here:

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