Answer:$ 506.05
Explanation:
Given
volume of container [tex] =20 m^3[/tex]
Let  L be the length of square-base and h be the height of Rectangular box
Cost of base[tex]=20 \$/m^2[/tex]
Cost of side and lid[tex]=10 \$ /m^2[/tex]
Cost of base [tex]c_1=L^2\times 20[/tex]
[tex]h=\frac{20}{L^2}[/tex]
cost of lid and side [tex]c_2=10\times 4L\cdot h+10\times L^2[/tex]
Total cost [tex]C=c_1+c_2[/tex]
[tex]C=20L^2+10L^2+40L\cdot h[/tex]
[tex]C=30L^2+\frac{800}{L}[/tex]
differentiate C w.r.t to L to get minimum cost
[tex]\frac{\mathrm{d} C}{\mathrm{d} L}=60L-\frac{800}{L^2}=0[/tex]
[tex]L^3=\frac{80}{6}[/tex]
[tex]L=2.37 m[/tex]
thus [tex]h=\frac{20}{5.613}=3.56 m[/tex]
Thus Lowest cost is [tex]C=30\times 5.617+\frac{800}{2.37}=$ 506.05[/tex]
