Question

The sum of surface areas of a cuboid with sides x 2 x x by 3 and a sphere is given to be constant prove that the sum of their volumes is minimum

Answer 1

dear frnd

For cuboidl=x3b=xh=2xSurface Area=2(lb+bh+hl)=2(x3.x+x.2x+2x.x3)=2(x23+2x2+2x23)=6x2Let radius of sphere be 'r'Surface Area=4πr2According to question:6x2+4πr2=C [Constant]  ⇒4πr2=C−6x2 ;(i)⇒r=C−6x2√2π√ ;(ii)Sum of volume=S=43πr3+l.b.h=43πr3+(x3)(x)(2x)=13[4πr3+2x3]Using  (ii), we get:S=13[4π(C−6x2√2π√)3+2x3]=13[4π(C−6x2√)8ππ√3+2x3]S=13⎡⎣(C−6x2)322π√+2x3⎤⎦Differentiating with respect to x, we get:3dSdx=ddx[(C−6x2)2π√32]+ddx(2x3)⇒3dSdx=12π√.32(C−6x2)32−1.ddx(C−6x2)+6x2⇒3dSdx=12π√.32(C−6x2)12(−12x)+6x2For maxima or minima dSdx=0⇒−12π√.32(C−6x2)12(12x)+6x2=0⇒1π√.3(C−6x2)12(12x)=24x2⇒3π√(C−6x2)12=2xSquaring both sides we get:9π(C−6x2)=4x2⇒C−6x2=4π9x2[Note: You have to check that d2SdS2<0 when C−6x2=4π9x2]Put this value in (i)4πr2=4π9x2⇒r2=x29⇒r=x3⇒x=3r [Hence proved]
if u like the answer then pls regard me as branliest