The solution ofV(x) = 0 in the domain ofVisx= 10.BecauseV(10) = 1000> V(1) = 149.5>V(10√3)= 0, this shows thatx= 10 maximizesVand that the maximum value ofVis 1000 cm184.108.40.206:Letxdenote the radius of the cylinder andyits height. Then its total surface area isπx2+2πxy=300π, sox2+ 2xy= 300. We are to maximize its volumeV=πx2y. Becausey=300−x22x,it follows thatV=V(x) =π2(300x−x3),0x10√3.It is then easy to show thatx= 10 maximizesV(x), thaty=x= 10 as well, and thus that the maximumpossible volume of the can is 1000πin.33.6.19:Letxbe the length of the edge of each of the twelve small squares. Then each of the three cross-shaped pieces will form boxes with base length 1−2xand heightx, so each of the three will have volumex(1−2x)2. Both of the two cubical boxes will have edgexand thus volumex3. So the total volume of allfive boxes will beV(x) = 3x(1−2x)2+ 2x3= 14x3−12x2+ 3x,0x12.NowV(x) = 42x2−24x+ 3;V(x) = 0 when 14x2−8x−1 = 0.The quadratic formula gives the twosolutionsx=114(4±√2). These are approximately 0.3867 and 0.1847, and both lie in the domain ofV.Finally,V(0) = 0,V(0.1847)≈0.2329,V(0.3867)≈0.1752, andV(0.5) = 0.25. Therefore, to maximizeV,one must cut each of the three large squares into four smaller squares of side length12each and form theresulting twelve squares into two cubes. At maximum volume there will be only two boxes, not five.3.6.20:Letxbe the length of each edge of the square base of the box and lethdenote its height. Then itsvolume isV=x2h. The total cost of the box is $144, hence4xh+x2+ 2x2= 144and thush=144−3x24x.ThereforeV=V(x) =x4(144−3x2)= 36x−34x3.172
The natural domain ofVis the open interval(0,4√3), but we may adjoin the endpoints as usual to obtaina closed interval. AlsoV(x) = 36−94x2,soV(x) always exists and is zero only atx= 4 (reject the other rootx=−4). Finally,V(x) = 0 at theendpoints of its domain, soV(4) = 96 (ft3) is the maximum volume of such a box. The dimensions of thelargest box are 4 ft square on the base by 6 ft high.3.6.21:Letxdenote the edge length of one square andythat of the other. Then 4x+4y= 80, soy= 20−x.The total area of the two squares isA=x2+y2, soA=A(x) =x2+ (20−x)2= 2x2−40x+ 400,with domain (0,20); adjoin the endpoints as usual. ThenA(x) = 4x−40, which always exists and whichvanishes whenx= 10. NowA(0) = 400 =A(20), whereasA(10) = 200. So to minimize the total area of thetwo squares, make two equal squares. To maximize it, make only one square.3.6.22:Letrbe the radius of the circle andxthe edge of the square.We are to maximize total areaA=πr2+x2given the side condition 2πr+ 4x= 100. From the last equation we infer thatx=100−2πr4=50−πr2.SoA=A(r) =πr2+14(50−πr)2=π+14π2r2−25πr+ 625for 0r50/π(becausex0). NowA(r) = 2π+14π2r−25π;A(r) = 0whenr=252 +π2=50π+ 4;that is, whenr≈7. Finally,A(0) = 625,A50π≈795.77andA50π+ 4≈350.06.