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Question-211145




Question Number 211145 by mr W last updated on 29/Aug/24
Commented by mr W last updated on 30/Aug/24
thanks! but i think people know   what i exactly mean.
thanks!butithinkpeopleknowwhatiexactlymean.
Commented by mr W last updated on 29/Aug/24
a corner of a solid cube with edge   length d is chopped off as shown.  find the radius R of the largest   solid sphere which can be cut from   the remaining object.  assume 0<a, b, c≤d.
acornerofasolidcubewithedgelengthdischoppedoffasshown.findtheradiusRofthelargestsolidspherewhichcanbecutfromtheremainingobject.assume0<a,b,cd.
Commented by MathematicalUser2357 last updated on 30/Aug/24
0<a,b,c≤d is not an equation or logic  you should say: “∧_(l∈{a,b,c}) (0<l∧l≤d)”
0<a,b,cdisnotanequationorlogicyoushouldsay:l{a,b,c}(0<lld)
Answered by mr W last updated on 04/Sep/24
say a, b, c are the x, y, x axes.  the equation of the cross section is  (x/a)+(y/b)+(z/c)=1  if no corner is chopped off, then  the largest sphere has  center at ((d/2),(d/2),(d/2)) and  radius R=R_0 =(d/2)  if  (((d/(2a))+(d/(2b))+(d/(2c))−1)/( (√((1/a^2 )+(1/b^2 )+(1/c^2 )))))≥R_0 =(d/2)  (1/a)+(1/b)+(1/c)−(√((1/a^2 )+(1/b^2 )+(1/c^2 )))≥(2/d)  R=R_0 =(d/2)  otherwise  case 1: center (R_1 , R_1 , d−R_1 ), radius R_1   (((R_1 /a)+(R_1 /b)+((d−R_1 )/c)−1)/( (√((1/a^2 )+(1/b^2 )+(1/c^2 )))))=R_1   ⇒(R_1 /d)=(((1/c)−(1/d))/( (√((1/a^2 )+(1/b^2 )+(1/c^2 )))−(1/a)−(1/b)+(1/c)))  case 2: center (d−R_2 , d−R_2 , R_2 ), radius R_2   ((((d−R_2 )/a)+((d−R_2 )/b)+(R_2 /c)−1)/( (√((1/a^2 )+(1/b^2 )+(1/c^2 )))))=R_2   ⇒(R_2 /d)=(((1/a)+(1/b)−(1/d))/( (√((1/a^2 )+(1/b^2 )+(1/c^2 )))+(1/a)+(1/b)−(1/c)))  case 3: center (d−R_3 , d−R_3 , d−R_3 ), radius R_3   ((((d−R_3 )/a)+((d−R_3 )/b)+((d−R_3 )/c)−1)/( (√((1/a^2 )+(1/b^2 )+(1/c^2 )))))=R_3   ⇒(R_3 /d)=(((1/a)+(1/b)+(1/c)−(1/d))/( (√((1/a^2 )+(1/b^2 )+(1/c^2 )))+(1/a)+(1/b)+(1/c)))  we can show that R_1 <R_3 , R_2 <R_3   that means the radius of maximum  sphere from case 1 to case 3 is R_3 .
saya,b,carethex,y,xaxes.theequationofthecrosssectionisxa+yb+zc=1ifnocornerischoppedoff,thenthelargestspherehascenterat(d2,d2,d2)andradiusR=R0=d2ifd2a+d2b+d2c11a2+1b2+1c2R0=d21a+1b+1c1a2+1b2+1c22dR=R0=d2otherwisecase1:center(R1,R1,dR1),radiusR1R1a+R1b+dR1c11a2+1b2+1c2=R1R1d=1c1d1a2+1b2+1c21a1b+1ccase2:center(dR2,dR2,R2),radiusR2dR2a+dR2b+R2c11a2+1b2+1c2=R2R2d=1a+1b1d1a2+1b2+1c2+1a+1b1ccase3:center(dR3,dR3,dR3),radiusR3dR3a+dR3b+dR3c11a2+1b2+1c2=R3R3d=1a+1b+1c1d1a2+1b2+1c2+1a+1b+1cwecanshowthatR1<R3,R2<R3thatmeanstheradiusofmaximumspherefromcase1tocase3isR3.

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