Question 1143682: A portion of a wire 70 inches in length is bent to form a rectangle having the greatest possible area such that length of the rectangle exceeds three times its width by 2 inches, and the dimensions of the rectangle are whole numbers. Find the length of the wire that is NOT used to form the rectangle.
So, I let w represent the width.
The length, therefore, would be 3w+2
And perimeter = 2L+2W
Therefore, I have 2(3w+2)+2w=70
Then 6W+4+2W=70
Then 8W+4=70
Then 8W=66
And finally, 66 divided by 8 is 8 with a remainder of 2.
So, is 2 inches the length of the wire that was not used? Is my work correct?
Answer by ikleyn(52785)   (Show Source):
You can put this solution on YOUR website! .
Your work and your final answer are both correct.
The only thing you forgot to mention, is THIS :
At given condition, to maximize the area of the rectangle, you should choose W as great as possible, satisfying other constraints
(the other constraint is to have integer length and width).
The rest is OK.
My congrats (!)
Good work (!)
Come again soon (!)
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