Two sides of a triangle are to be 10 cm long. How long should the third side be to make the triangle have maximum area? (Hint: square the area then differentiate to find the value x for which the area has maximum value)
So the triangle must be isosceles.
Sketch it that way, draw a perpendicular to the missing side.
let that perpendicular, the height, be h cm
let the base be 2x cm (I am avoiding fractions)
A = (1/2)(2x)(h) = xh
then A^2 = x^2h^2, (their hint)
but x^2 + h^2 = 10^2
h^2 = 100 - x^2
A^2 = x^2(100-x^2) = 100x^2 - x^4
d(A^2)/dx = 200x - 4x^3 = 0 for a max/min
200x - 4x^3 = 0
x^3 - 50x = 0
x(x^2 - 50) = 0
x = 0 <----- that would yield the minimum area
or
x^2 = 50
x = 5√2
but I called the base 2x, so the third side must be 10√2 cm
(notice this would make the base angle 45°, and the vertex angle 90°
that is, you have half a square.
This makes sense since the largest area of a rectangle is yielded
when that rectangle is a square )
To find the length of the third side of the triangle that maximizes the area, we can use the concept of derivatives.
Let's assume the length of the third side is denoted as 'x'. We know that the two sides given are 10 cm long. Now, we can use the formula for the area of a triangle, which is given by:
Area = (1/2) * base * height
In this case, the base of the triangle is x, and the height can be determined using the two sides given.
Using the given information, we can calculate the height of the triangle using the Pythagorean theorem. Let's call it 'h':
h = √(10^2 - x^2)
Now, substitute the base and height values into the area formula:
Area = (1/2) * x * √(10^2 - x^2)
To find the value of 'x' that maximizes the area, we need to differentiate the area formula with respect to 'x'.
d/dx (Area) = d/dx [(1/2) * x * √(100 - x^2)]
Now, square the area formula to get rid of the square root:
Area^2 = [(1/2) * x * √(100 - x^2)]^2
Differentiate this squared area formula:
d/dx (Area^2) = 2 * [(1/2) * x * √(100 - x^2)] * ( √(100 - x^2) * d/dx(x) + x * d/dx(√(100 - x^2)))
Simplifying this expression, we get:
d/dx (Area^2) = x * √(100 - x^2) + x * (√(100 - x^2)) * (-x/√(100 - x^2))
Simplifying further:
d/dx (Area^2) = x * √(100 - x^2) - (x^2/√(100 - x^2))
To find the value of 'x' at which the area is maximized, we need to set the derivative to zero and solve for 'x':
x * √(100 - x^2) - (x^2/√(100 - x^2)) = 0
x * √(100 - x^2) = x^2/√(100 - x^2)
Multiplying both sides by √(100 - x^2), we get:
x * (√(100 - x^2))^2 = x^2
x * (100 - x^2) = x^2
100x - x^3 = x^2
Rearranging terms, we get a cubic equation:
x^3 + x^2 - 100x = 0
To solve this cubic equation, we can use numerical methods or a graphing calculator to find the value of 'x' that maximizes the area of the triangle.