Answer:
[tex]2 ft^3[/tex]
Step-by-step explanation:
We are given that
Side of square shaped piece= 3ft
Let square of side x cut from each corner of square piece cardboard.
Length of box=[tex]3-2x[/tex]
Breadth of box=[tex]3-2x[/tex]
Height of box= x
We have to find the largest volume of box.
Volume of box is given by
[tex]V=x(3-2x)^2[/tex]
[tex]V=(4x^3-12x^2+9x)[/tex]
Differentiate w.r.t x
[tex]\frac{dV}{dx}=12x^2-24x+9[/tex]
Substitute [tex]\frac{dV}{dx}=0[/tex]
[tex]12x^2-24x+9=0[/tex]
[tex]4x^2-8x+3=0[/tex]
[tex](2x-1)(2x-3)=0[/tex]
[tex]2x-1=0\implies x=\frac{1}{2}[/tex]
[tex]2x-3=0\implies x=\frac{3}{2}[/tex]
[tex]x=\frac{1}{2}, x=\frac{3}{2}[/tex]
Again differentiate w.r.t x
[tex]\frac{d^2V}{dx^2}=24x-24[/tex]
Substitute [tex]x=\frac{1}{2}[/tex]
[tex]\frac{d^2V}{dx^2}=12-24=-12<0[/tex]
Substitute [tex]x=\frac{3}{2}[/tex]
[tex]\frac{d^2V}{dx^2}=36-24=12>0[/tex]
Therefore, [tex]\frac{d^2V}{dx^2} <0[/tex] at [tex]x=\frac{1}{2}[/tex]
Hence, the volume is maximum at [tex]x=\frac{1}{2}[/tex]
Substitute the value [tex]x=\frac{1}{2}[/tex] then we get
[tex]V=\frac{1}{2}(3-1)^2=2 ft^3[/tex]
