SOLUTION: Can someone please help? Find the dimensions of the box described. The length is twice as long as the width. The height is 4 inches greater than the width. The volume is 48 cu

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Question 1137042: Can someone please help?
Find the dimensions of the box described.
The length is twice as long as the width. The height is 4 inches greater than the width. The volume is 48 cubic inches.
length:
width:
height:

Answer by ikleyn(52855)   (Show Source): You can put this solution on YOUR website!
.
Let x be the width.


Then the length is 2x  and the height is (x+4).


The volume equation is


    x*(2x)*(x+4) = 48.


    2x^2*(x+4) = 48

    x^2*(x+4) = 24


By the "trial and error" method, x= 2.



The plot below shows that the guessed solution x = 2 is UNIQUE.



    


    Plot y = x*(2x)*(x+4) (red) and y = 48 (green)


Also, notice that the function  f(x) = x*(2x)*(x+4) is monotonic over positive x, so the uniqueness of the solution 
is OBVIOUS even without the plot.

  
ANSWER.  The width is 2 in;  the length is  2*2 = 4 in;  the height is (2+4) = 6 in.


CHECK.    2*4*6 = 48.    ! Correct !


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