【導(dǎo)讀】[解析]設(shè)圓柱的底面半徑為r,高為h,體積為V,則4r+2h=l,則V′=lπr-6πr2,令V′=0,得r=0或r=l6,而r>0,當(dāng)r=l6時(shí),V取得最大值,最大值為3π.圓柱的側(cè)面積S=2πx·h,∴S2=16π2,′=16π2,由′=0,得x=22r,∴Smax. V=12x2·sin60°·l,∴l(xiāng)=4V3x2,S′表=3-43Vx2=0,∴x3=4V,即x=34V.-x3900+400x,0≤x≤390,90090-100x-20200,x>390,由P′=0,得x=300,故選D.∴函數(shù)y=(x-1)4的單調(diào)遞增區(qū)間為,當(dāng)a≤0時(shí)y′≥0恒成立,∴函數(shù)y=x3-3ax+6在上是增函數(shù),∴函數(shù)y=x3-3ax+6在上是減函數(shù),故應(yīng)選A.由ex+a=0得ex=-a,∴x=ln(-a),故當(dāng)矩形場(chǎng)地的長(zhǎng)為4m,寬為4m時(shí),面積取最大值16m2.V′=12x2-384x+2304=12,[解析]要求材料最省就是要求新砌的墻壁總長(zhǎng)度最短,如下圖所示,設(shè)場(chǎng)地寬為x米,則長(zhǎng)為512x米,因此新墻壁總長(zhǎng)度L=2x+512x(x>0),則L′=2-512x2.當(dāng)x=16時(shí),L極小值=Lmin=64,[解析]利潤(rùn)為S==-x2+230x-6000,