Answer:
[tex]\frac{5}{3}[/tex] in
Step-by-step explanation:
Let x be the side of square.
Length of box=8-2x
Width of box=15-2x
Height of box=x
Volume of box=[tex]L\times B\times H[/tex]
Substitute the values then we get
Volume of box=V(x)=[tex](8-2x)(15-2x)x=(15x-2x^2)(8-2x)[/tex]
[tex]V(x)=120x-30x^2-16x^2+4x^3[/tex]
[tex]V(x)=4x^3-46x^2+120x[/tex]
Differentiate w.r.t x
[tex]V'(x)=12x^2-92x+120[/tex]
[tex]V'(x)=0[/tex]
[tex]12x^2-92x+120=0[/tex]
[tex]3x^2-23x+30=0[/tex]
[tex]3x^2-18x-5x+30=0[/tex]
[tex]3x(x-6)-5(x-6)=0[/tex]
[tex](x-6)(3x-5)=0[/tex]
[tex]x-6=0\implies x=6[/tex]
[tex]3x-5=0\implies x=\frac{5}{3}[/tex]
Again differentiate w.r.t x
[tex]V''(x)=24x-92[/tex]
Substitute x=6
[tex]V''(6)=24(6)-92=52>0[/tex]
Substitute x=5/3
[tex]V''(5/3)=24(5/3)-92=-52<0[/tex]
Hence, the volume is maximum at x=[tex]\frac{5}{3}[/tex]
Therefore, the side of the square ,[tex]x=\frac{5}{3}[/tex] in cutout that gives the box the largest possible volume.
