area of a rectangle

Question

The area of a rectangle is x⁴-y⁴/8xy and the length is x²-y²/2xy

a. What is the width of the rectangle?

b. If you double the length of the rectangle but keep the same width, what is the new area of the rectangle?

Answer 1

It appears you have left off grouping parenthesis.... THESE ARE IMPORTANT!!

Note:  Area = Length*width ... A=Lw   ... w = A/L

 

A = (x⁴-y⁴)/(8xy)   and    L = (x²-y²)/(2xy)    

 

W = [(x⁴-y⁴)/(8xy)]/[(x²-y²)/(2xy)]

 

w = [(x⁴-y⁴)/(8xy)][(2xy)/(x²-y²)]

 

Note:  we can cancel the 2xy in numerator of 2nd term into the 8xy in denominator of

1st term, leaving 4 in denominator of 1st term.

 

w = [(x⁴-y⁴)/4][1/(x²-y²)]

 

Note that we can use the formula for the difference of two squares here:

a² - b² = (a-b)(a+b)  and thus   x⁴-y⁴ = (x²)² - (y²)² = (x²-y²)(x²+y²)

 

w = [(x²-y²)(x²+y²)/4][1/(x²-y²)]

 

We can now cancel the (x²-y²) terms from the numerator and denominator terms

 

w = (x²+y²)/4

 

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As for the 2nd part of the question. 

Since area = length*width ...  A = Lw

 

If we double the length and keep the width the same we have

 

A = 2Lw  so all it does is double the area

 

Thus the new area is the old area multiplied by 2

 

A = 2[(x⁴-y⁴)/(8xy)] = (x⁴-y⁴)/(4xy)