Home › JEE Main › Mathematics › Application of Derivatives › Maximum area of a rectangle inscribed in a semic…
Maximum area of a rectangle inscribed in a semicircle of radius R (base on diameter):
AR²
BπR²
CR²/2
D2 R² / sqrt(2) = R² sqrt(2)
Answer & Solution
Correct answer: D. 2 R² / sqrt(2) = R² sqrt(2)
Width = 2x, height = sqrt(R² - x²). Area A = 2x sqrt(R² - x²). dA/dx = 0 gives x = R/sqrt(2). A_max = 2(R/sqrt(2)) sqrt(R² - R²/2) = sqrt(2) R × (R/sqrt(2)) × sqrt(2) = R² sqrt(2). Wait, recompute: A = 2 × (R/√2) × √(R²/2) = 2 × (R/√2) × (R/√2) = R². So max area = R². (Width = R√2, height = R/√2.)
Related questions
Elasticity of demand is given byProfit is maximised whereMarginal revenue (MR) for a price-taking firm (perfect competition) equalsMarginal cost (MC) isTwo numbers have a sum of $24$. Their product is largest when the numbers areFor $f(x)=3x^4-8x^3+12x^2-48x+25$ on $[0,3]$, the critical point inside the interval isA cylindrical tank of radius $10$ m is filled with wheat at $314$ m$^3$/h. The depth of thThe maximum value of $[x(x-1)+1]^{1/3}$ for $0\le x\le 1$ is