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Question Number 132905 by bramlexs22 last updated on 17/Feb/21
Answered by bobhans last updated on 17/Feb/21
Totalsurfacearea=A=πrh+2hr+πr2=200cm2h=200−πr22r+πrVolofblock=V=12πr2hV=12πr2(200−πr22r+πr)=πr2(200−πr22+π)(a)V=πr(200−πr2)4+2π
(b)maximumvalueofV.dVdr=12π+4ddr[200πr−π2r3]=0⇔12π+4[200π−3π2r2]=0⇒r=2003πVmax=π(2003π)(200−π(2003π))2π+4Vmax=4000π2(2π+4)3π=20002π(π+2)3cm3
Commented by bramlexs22 last updated on 17/Feb/21
thanks
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