For a given perimeter,the triangle having the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the perimeter of the triangle be $P$. According to Heron's formula,the area $A$ of a triangle with sides $a, b, c$ is given by $A = \sqrt{s(s-

Vedclass · Accepted Answer

Equilateral triangle

Vedclass · Answer

(C) Let the perimeter of the triangle be $P$. According to Heron's formula,the area $A$ of a triangle with sides $a, b, c$ is given by $A = \sqrt{s(s-a)(s-b)(s-c)}$,where $s = P/2$ is the semi-perimeter.For a fixed $s$,the product $(s-a)(s-b)(s-c)$ is maximized when $s-a = s-b = s-c$,which implies $a = b = c$.Thus,for a given perimeter,the triangle with the maximum area is an equilateral triangle.Alternatively,using the method of Lagrange multipliers or calculus,we find that for a fixed perimeter,the area is maximized when all sides are equal.

For a given perimeter,the triangle having the maximum area is:

Similar Questions

Find the absolute maximum value and the absolute minimum value of the function given by $f(x) = 4x - \frac{1}{2}x^2$ for $x \in \left[-2, \frac{9}{2}\right]$.

If $m$ and $M$ respectively denote the minimum and maximum of $f(x)=(x-1)^2+3$ for $x \in [-3, 1]$, then the ordered pair $(m, M)$ is equal to

Let $f(x) = (x^2 - 1)^n (x^2 + x + 1)$. Then $f(x)$ has a local extremum at $x = 1$ when:

Find the local maximum and local minimum values for the function $h(x) = \sin x + \cos x$ in the interval $0 < x < \frac{\pi}{2}$.

The maximum slope of the curve $y = \frac{1}{2} x^{4} - 5 x^{3} + 18 x^{2} - 19 x$ occurs at the point

Vedclass Test Series

Exam Paper Generator

Online Exam Module