Find the range of possible values for the width of the base so that the volume of the figure will lie between 2 cm^3 and 4cm^3.A rectangular solid...

length of the base =
2times

Let length and width of the base be 2x and x
.

Let h be the height.

Then
the sum of the length of all the edges of the rectangular solis = 4(length+width+height)
= 40.

4(2x+x+h) = 40.

3x+h =
10.

Therefore h =
10-3x.

Volume of the given solid= length*width*height =
2x*x(10-3x) should be btween 2 to 4  cm^3.

2x^2(10-3x) = 2
to 4 or

x^2(10-3x) = 1 to 2
cm^3.

x = 0.34 to to 0.48.

At
the lowest volume x = 0.34 (2 decimal accuracy)

Therefore
width x = 0.34  length = 2x = 0.76 and h = 10-3x = 8.98. Then total length =
4(0.34*3+8.98) = 40cm

Volume = 0.34*0.68*8.98 = 2.076
cm^3.

The height volume x = 0.48 (2 decimal point
accuracy):

(Width , length, height) = (0.48 , 2*48 ,
10-3*0.48) = (0.48 , 0.96 ,  8.56). Total wire length =40cm and volume = 0.48*0.96*8.56
= 3.944 cm^3.

So  values  of the   width  w  length l and
height h are given by:

(w, l , h) = {x , 2x , (10-3x)} , x
belong to the interval (0.34 , 0.48).