• Perimeter of a triangle is 8 cm.
• One of the sides of the triangle is 3 cm.
• The area of the triangle is maximum.
Let us consider,
• ‘x’ and ‘y’ be the other two sides of the triangle.
Now, perimeter of the ΔABC is
8 = 3 + x + y
y = 8-3-x = 5-x
y = 5-x --- (1)
Consider the Heron’s area of the triangle,
As perimeter = a + b+ c = 8
Now Area of the triangle is given by
Now substituting (1) in the area of the triangle,
[squaring on both sides]
Z = A2 = 4(5x –x2-4) ----- (2)
For finding the maximum/ minimum of given function, we can find it by differentiating it with x and then equating it to zero. This is because if the function f(x) has a maximum/minimum at a point c then f’(c) = 0.
Differentiating the equation (2) with respect to x:
To find the critical point, we need to equate equation (3) to zero.
20 – 8x = 0
8x = 20
Now to check if this critical point will determine the maximum area of the triangle, we need to check with second differential which needs to be negative.
Consider differentiating the equation (3) with x:
As , so the function A is maximum at .
Now substituting in equation (1):
y = 5 – 2.5
y = 2.5
As x = y = 2.5, two sides of the triangle are equal,
Hence the given triangle is an isosceles triangle with two sides equal to 2.5 cm and the third side equal to 3cm.
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