91- 6303239042
Maximum Area Under a Fixed Perimeter
For a rectangle with sides $$x$$ and $$y$$ and fixed perimeter $$p$$:
$$x+y=\frac{P}{2}$$
The area is maximum when:
$$x=y=\frac{P}{4}$$
Usage
- Used in rectangular area optimization problems.
Sign in
Please select an account to continue using cracku.in
↓ →
Maximum Area Under a Fixed Perimeter
For a rectangle with sides $$x$$ and $$y$$ and fixed perimeter $$p$$:
$$x+y=\frac{P}{2}$$
The area is maximum when:
$$x=y=\frac{P}{4}$$
Usage
- Used in rectangular area optimization problems.
Start your IIM journey with the right preparation and crack CAT 2026.
Educational materials for CAT preparation
