Question Number 101680 by john santu last updated on 04/Jul/20

Answered by bemath last updated on 04/Jul/20

let point R(x,x^2 ) area of ΔPQR =(1/2) determinant ((( 2 4 1)),((−1 1 1)),(( x x^2 1))) A(x) =(1/2){2(1−x^2 )−4(−1−x)+1(−x^2 −x)} A(x)=(1/2){2−2x^2 +4+4x−x^2 −x} A(x)= (1/2){−3x^2 +3x+6} ⇔((dA(x))/dx) = (1/2){−6x+3} = 0 , x = (1/2) ⇔ ((d^2 A(x))/dx^2 ) = −6 < 0 ⇐ has max value area triangle PQR has maximum where the point R((1/2),(1/4)) then maximum area is (1/2){((−3)/4) +(3/2)+6} = (1/2){6+(3/4)}= ((27)/8) ★

Commented byjohn santu last updated on 04/Jul/20

correct .

Answered by mr W last updated on 04/Jul/20

R(x,y) tangent at R must be // PQ such that Δ_(PQR) has maximum area. y′=2x=((4−1)/(2+1))=1 ⇒x=(1/2) ⇒y=(1/4) max. Δ_(PQR) =∣(1/2) determinant (((2+1),(4−1)),(((1/2)+1),((1/4)−1)))∣ =∣(1/2)(−(9/4)−(9/2))∣=((27)/8)

Commented bybemath last updated on 04/Jul/20

sir how to proof tangent at R must be paralell to PQ?

Commented bymr W last updated on 04/Jul/20

base PQ is given. maximum area means then maximum altitude from R to base PQ, i.e. R must have the largest distance to PQ. this is the case when the line through R and parallel to PQ tangents the curve.

Commented bymr W last updated on 04/Jul/20

Commented bybemath last updated on 04/Jul/20

ok sir. i understand. thank you

Commented byjohn santu last updated on 04/Jul/20

great