the dimensions that maximize its volume is
Length and width (x)=4√33 m
Height (h)=2√33m
Volume of rectangular box is defined as the product of the area of square base and height of box.
Formula for volume is
V=x2.h
Where x be the length and width of rectangular box and h is height.
Formula used for area of box is
A=x2+4xh
We can use second derivative test for checking the volume is maximum.
If
d2Vdx2<0 volume is maximum.
To find dimensions of rectangular box
Let x be the length and width of square base box. h be the height of box.
Area of box is
A=x2+4xh=16
⟹4xh=16−x2
h=16−x2 / 4x
Volume of box
V=x2h
V=x2(16−x2 / 4x)
V=4x−x3 / 4
Taking first derivative of V w.r.t. x
dV / dx=4−3x2 / 4
Putting the derivative of V is equal to zero and we get value of x
dV / dx=0
⟹4−3x2 / 4=0
⟹3x2=16
x2=163
x2=√163
x=4√3
x=4√3 / 3
Substitute the value of x in h and we get value of h
Taking first derivative of V w.r.t. x
dV / dx=4−3x2 / 4
Putting the derivative of V is equal to zero and we get value of x
dV / dx=0
⟹4−3x2 / 4=0
⟹3x2=16x2=163x2=√163
x=4√3
x=4√3 / 3
Substitute the value of x in h and we get value of h
h=16−x2 / 4x
⟹h=16−(4/√3)2 / 4×4/√3
h=16−16/3 / 16/√3
h=2√3 / 3m
Taking second derivative of V w.r.t. x and maximize at value of x = 14 d2V / dx2=−3x / 2
d2Vdx2=−3(4√3/3)/2
=−2√3<0
d2Vdx2<0
which is maximum Dimensions of rectangular box are
Length and width (x)=4√3 / 3 m
Height (h)=2√3 / 3m
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