Determine the volume of the largest box in the first octant with three in the coordinate planes, one vertex at the origin, and its opposite vertex in the plane x+3y+5z=15

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The volume of the largest rectangular box is to be determined based on the first octant with three faces in the coordinate planes, and one vertex in the plane x+3y+5z=15. The volume of the rectangular box that lies in the first octant based on the three faces that lies in the coordinate plane as follows, V=f(x,y)=xyz The vertex get lies in the plane as below, x+3y+5z=15 5z=15–x–3y z=15–x–3y5 The volume is determined as below, V=f(x,y)=15xy(15−x−3y)=3xy−x2y5−3xy25 The Volume will be maximum if fx=fy=0 As f(x)=3y−25xy−35y2=15y(15−2x−3y)=0 y=0, y=15−2x3 (1) And fy=3x−x25−6xy5=15x(15−x−6y)=0 x=0 and substituting y=0 x=15 At y=15−2x3 fy=0 15x(15−x−6⋅15−2x3)=0 x=0,15−x−30+4x=0 3x=15 x=5 y=15−2⋅03=153 At x=0 y=5 And at x=5, y=53 The Critical point are (0,0),(15,0),(0,5),(5,53) and z At (5,53)isz=15−x−3y5 z=15−5−55 z=1 The Volume of the longest rectangular box is determined as below, 5⋅53=253 Hence, the volume is 253

Determine the volume of the largest box in the first octant with three in the coordinate planes, one vertex at the origin, and its opposite vertex in the plane x+3y+5z=15

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