Solve the problem.An open box is to be formed from a rectangular sheet of tin 20*32 in cutting equal squares , x in on a side , from the four cornes and turning up the sides . Express the volume...

Solve the problem.

An open box is to be formed from a rectangular sheet of tin 20*32 in cutting equal squares , x in on a side , from the four cornes and turning up the sides .

Express the volume of the box as a function of x .

Asked on by maisaphie

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Top Answer

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

A rectangular shape has the volume of:

V = A*h

A-the area of the base

h-height of the box

The area of the base is the area of a rectangle, whose width and length are:

w = 20 - 2x

l = 32 - 2x

A = w*l

A = (20 - 2x)(32 - 2x)

We'll factorize by 2 both pair of brackets:

A = 2*2(10 - x)(16 - x)

A = 4(10 - x)(16 - x)

The height of the box is x.

We'll calculate the volume:

V = 4x(10 - x)(16 - x)

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

We know that the open box has 5 faces.

Each face is a square.

Therefore 20.32 is  divided into 5 equal areas.

So each face has the area = 20*32/5 = 128 sq in

Therefore each side of the square face should be sqrt128.

Therefore the volume of the box = (sqrt128)^3 = sqrt(128^3) =1448.1547 cubic in

nunou1's profile pic

nunou1 | Student, Grade 9 | eNotes Newbie

Posted on

Area of base=length.width=(30-2X).(20-2X)=4Xsquare-100X+600 inchsquare.height=X.

Therefor volume=area of base times X=

(4Xsquare-100X+600)timesX=4Xcube-100Xsquare+600X

william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

The sides of the rectangular sheet are 20 in. and 32 in. in length.

Let the sides of the square which are cut be x.

Once they are cut the side that was initially 20 in. long is now 20- 2x in. long. And the side that was 32 in. long remains 32-2x in. long.

Now the sides are turned up to make a box.

The base of the box has dimensions 20 - 2x and 32- 2x in. And the height of the box is x in.

Therefore the volume of the box is (20- 2x)(32-2x)(x)

Or expressed as a function of x :

V(x) = 4x(10- x)(16-x) in^3

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