Find the area (in sq. units) of the largest rectangle with lower base on the x-axis upper vertices on thecurve mathrmy=12- mathrmx2.

Question

Find the area (in sq. units) of the largest rectangle with lower base on the x-axis upper vertices on thecurve mathrmy=12- mathrmx2. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

A=2 x y f prime(x)=A(2 x)(12-x2) f prime(x)=2[12-3 x2] x=2 f prime prime(x)=2[-6]= max y=12-4=8 text Area =2 × 8 × 2=32 text sq. units

Q. Find the area (in sq. units) of the largest rectangle with lower base on the x-axis \& upper vertices on thecurve $y = 12 - x^{2}$ .

$A = 2 x y$
$f^{'} (x) = A (2 x) (12 - x^{2})$
$f^{'} (x) = 2 [12 - 3 x^{2}]$
$x = 2$
$f^{''} (x) = 2 [- 6] = max$
$y = 12 - 4 = 8$
$Area = 2 \times 8 \times 2 = 32 sq. units$

Q. Find the area (in sq. units) of the largest rectangle with lower base on the x-axis \& upper vertices on thecurve y=12−x2.

A=2xy f′(x)=A(2x)(12−x2) f′(x)=2[12−3x2] x=2 f′′(x)=2[−6]=max y=12−4=8 Area =2×8×2=32 sq. units

Q. Find the area (in sq. units) of the largest rectangle with lower base on the x-axis \& upper vertices on thecurve $y = 12 - x^{2}$ .

$A = 2 x y$
$f^{'} (x) = A (2 x) (12 - x^{2})$
$f^{'} (x) = 2 [12 - 3 x^{2}]$
$x = 2$
$f^{''} (x) = 2 [- 6] = max$
$y = 12 - 4 = 8$
$Area = 2 \times 8 \times 2 = 32 sq. units$