+0  
 
+1
2
3
avatar+73 

Square A and Square B are both 2009 by 2009 squares. Square A has both its length and width increased by an amount x, while Square B has its length and width decreased by the same amount x. What is the minimum value of x such that the difference in area between the two new squares is at least as great as the area of a 2009 by 2009 square?

 

Thanks so much!

 Jun 19, 2024

Best Answer 

 #2
avatar+73 
+1

The new area of Square A is (2009 +x)^2, while the new area of Square B is (2009 - x)^2. The difference is \(\begin{align*} &(2009+x)^2-(2009-x)^2\\ &\qquad=(2009+x+2009-x)(2009+x-2009+x) \\ &\qquad=(2\cdot 2009)(2x) \end{align*}\)

For this to be at least as great as the area of a 2009 by 2009 square, \(2(2009)2(x)\geq 2009^2\Rightarrow x\geq \boxed{\frac{2009}{4}}.\)

 Jun 19, 2024
 #1
avatar+506 
-1

Let's analyze the areas of the new squares and set up an inequality to find the minimum value of x.

 

Original Areas:

 

Area of Square A: 2009 * 2009 (square units)

 

Area of Square B: 2009 * 2009 (square units)

 

New Areas:

 

New area of Square A: (2009 + x) * (2009 + x) (square units)

 

New area of Square B: (2009 - x) * (2009 - x) (square units)

 

Difference in Area:

 

We want the difference in area between the new squares to be at least as great as the area of a 2009 by 2009 square. So, we can set up an inequality:

 

(New area of Square A) - (New area of Square B) ≥ Area of a 2009 by 2009 square

 

Expanding the Inequality:

 

[(2009 + x) * (2009 + x)] - [(2009 - x) * (2009 - x)] ≥ 2009 * 2009

 

Simplifying the Inequality:

 

Expand both squares and subtract like terms:

 

4x^2 ≥ 2009 * 2009

 

Solving for x:

 

Divide both sides by 4:

 

x^2 ≥ (2009 * 2009) / 4

 

Take the square root of both sides (remembering that x can be positive or negative since the area cannot be negative):

 

x ≥ ± (2009 / 2)

 

Finding the Minimum Value:

 

We want the minimum non-negative value of x. Since the area cannot be negative, we discard the negative solution. Therefore:

 

x ≥ 2009 / 2

 Jun 19, 2024
 #3
avatar+73 
0

That's incorrect. Thanks, tho!

PurpleWasp  Jun 19, 2024
 #2
avatar+73 
+1
Best Answer

The new area of Square A is (2009 +x)^2, while the new area of Square B is (2009 - x)^2. The difference is \(\begin{align*} &(2009+x)^2-(2009-x)^2\\ &\qquad=(2009+x+2009-x)(2009+x-2009+x) \\ &\qquad=(2\cdot 2009)(2x) \end{align*}\)

For this to be at least as great as the area of a 2009 by 2009 square, \(2(2009)2(x)\geq 2009^2\Rightarrow x\geq \boxed{\frac{2009}{4}}.\)

PurpleWasp Jun 19, 2024

1 Online Users

avatar