You must design a closed rectangular box of width w, length l and height h, whose volume is 504 cm3. The sides of the box cost 3 cents/cm2 and the top and bottom cost 4 cents/cm2. Find the dimensions of the box that minimize the total cost of the materials used.

Answer:Dimensions will be Length = 7.23 cmWidth = 7.23 cmHeight = 9.64 cmStep-by-step explanation:A closed box has length = l cmwidth of the box = w cmheight of the box = h cmVolume of the rectangular box = lwh 504 = lwh[tex]h=\frac{504}{lw}[/tex]Sides which involve length and width and height, cost = 3 cents per cm²Top and bottom of the box costs = 4 cents per cm²Cost of the sides [tex]C_{s}[/tex]= 3[2(l + w)h] = 6(l + w)h[tex]C_{s}[/tex]= 3[2(l + w)h][tex]C_{s}=6(l+w)(\frac{504}{lw} )[/tex]Cost of the top and the bottom [tex]C_{(t,p)}[/tex]= 4(2lw) = 8lwTotal cost of the box C = [tex]3024\frac{(l+w)}{lw}[/tex] + 8lw                                       = [tex]3024[\frac{1}{l}+\frac{1}{w}][/tex] + 8lwTo minimize the cost of the sides [tex]\frac{dC}{dl}=3024(-l^{-2}+0)+8w=0[/tex][tex]\frac{3024}{l^{2}}=8w[/tex][tex]\frac{378}{l^{2}}=w[/tex] ---------(1)[tex]\frac{dC}{dw}=3024(-w^{-2})+8l=0[/tex][tex]\frac{3024}{w^{2}}=8l[/tex][tex]\frac{378}{w^{2}}=l[/tex] [tex]w^{2}=\frac{378}{l}[/tex]-------(2)Now place the value of w from equation (1) to equation (2)[tex](\frac{378}{l^{2}})^{2}=\frac{378}{l}[/tex][tex]\frac{(378)^{2} }{l^{4}}=\frac{378}{l}[/tex]l³ = 378l = ∛378 = 7.23 cmFrom equation (2) [tex]w^{2}=\frac{378}{7.23}[/tex][tex]w^{2}=52.28[/tex]w = 7.23 cmAs lwh = 504 cm³(7.23)²h = 504[tex]h=\frac{504}{(7.23)^{2}}[/tex]h = 9.64 cm