# 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 is to be constructed with a special kind of wire...

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 is to be constructed with a special kind of wire along all the edges. The length of the base is to be twice of the width of ...the base. The height of the rectangular solid is such that the total amount of wire used (for the whole figure) is 40 cm.

Write your answer correct to two decimal places.

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The solid that you want to build has a rectangular base with length equal to twice the width. Let the width be represented as W, so we have the length as 2W. Let the height of the solid be H.

Now we have the total amount of wire needed to construct this as:

4*(2W + W + H) which is equal to 40.

So 2W+ W+ H = 10

=> 3W + H = 10.

=> H = 10 - 3W

Now volume is equal to 2W*W*(10 - 3W) = 20*W^2- 6*W^3 and it lies between 2 and 4.

=> 10 W^2 - 3 W^3 lies between 1 and 2

=> 1 < 10 W^2 - 3 W^3 < 2

=> 1 < W^2 (10 -3W) < 2

Now we can only use trial and error and arrive at the result. You could use Excel to feed in the relation and use the Goal Seek function. We get W= .34 for the expression to be greater than 1 and W= .48 for the expression to be less than 2.

Therefore the width should be greater than .34 and less than .48

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).