CUET Preparation Today
A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. Two sides having fence are of same length x. The maximum area enclosed by the park is: |
$\frac{1}{2}x^2$ $\pi x^2$ $\frac{3}{2}x^2$ $\sqrt{\frac{x^3}{8}}$ |
$\frac{1}{2}x^2$ |
Area of the park, $A=\frac{1}{2}x^2\sin θ$. Now A has maximum when $θ=π/2$ ∴ Maximum area = $x^2/2$ |
