Question 125564This question is from textbook Algebra Structure and Method
: I have been struggling on this word problem for a while now and I was wondering if anyone could help me? I would deeply appreciate it!!! Please and Thank You!!
Solving Problems Involving Inequalities
Solve.
The length of a rectangle exceeds the width by 10cm. If each dimension were increased by 3cm, the area would be no less than 111cm^2 more. What are the least possible dimensions of the rectangle?
This question is from textbook Algebra Structure and Method
Found 2 solutions by solver91311, stanbon:
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website! Solve this as if it were stated that the increase in area were equal to 111 cm^2. Then adjust the expressions for the dimensions to make them appropriate inequalities.
Original width of the rectangle: x
Original length of the rectangle: x + 10
Area of the original rectangle: 
New width: x + 3
New length: x + 10 + 3 = x + 13
Area of new rectangle: 
But  , so by substitution: 
Expand, put all the terms with variables on the left leaving the constants on the right, and collect terms:
 , so we have 12 as a length of the original rectangle IF we consider the increase in area to be exactly 111 sq cm. That implies that the length of the original rectangle would have been 22.
 and  which is 111 larger, so that checks so far. But now we have to deal with the fact that the problem wants the area increase to be no less than 111, in other words  .
We know that 'or equal' is going to be part of our expressions for the original width and length, but which way will the rest of the inequality go? The easiest thing to do is pick something for the width that is smaller (or larger) than 12 and see what happens. Let's choose 11.
If the width is 11, then the length is 21, so the area of the original rectangle would be 231. The new rectangle would have dimensions of 14 and 24, so the new area would be 336, and the difference is 105. That is less than 111, so to say  and  would be incorrect. The correct expressions are  and  .
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! The length of a rectangle exceeds the width by 10cm.
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Original dimensions and area:
width = x cm ; length = x+10 cm ; Area = x(x+10) = x^2+10x cm^2
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If each dimension were increased by 3cm, the area would be no less than 111cm^2 more.
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New dimensions and area:
width= x+3 cm ; length= x+13 cm ; Area = (x+3)(x+13) cm^2
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What are the least possible dimensions of the rectangle?
Inequalities:
(x+3)(x+13) >=111
x^2+16x+39-111>=0
x^2+16x-72 >= 0
Graph the parabola to see where
it is >= 0 and x is positive:
x >= 3.661904...
Dimensions: width = 6.661904... ; length = 16.661904...
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Cheers,
Stan H.
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