GMAT Club World Cup 2022 (DAY 2): A green rectangle is inscribed in

Let length of rectangle = L, width of rectangle = W
Required area (RA) = Area of entire right triangle + area of green rectangle
RA = {1/2 * (12+W) * (5+L)} + {L*W}
To minimize RA, we need minimum values for L and W

Let us test some values: (Note: Values can be same as well since a rectangle can be a square as well)
L=1,W=1: RA=39+1=40
L=1,W=2: RA=42+2=44
L=2,W=1: RA=45.2+2=47.5
L=2,W=2: RA=49+4=53
L=2,W=3: RA=52.5+6=58.5
L=3,W=2: RA=56+6=62

So, we can see that either area can be less than or greater than 60 but cannot be equal to 60

If we consider: L=5,W=3, then RA=75+15=90

So minimum possible RA (required area) = 90

Answer - B

As we have similar triangles, xy=60
2xy=120
Hence, options A, B and C are ruled out.
Area of 2 small similar triangles+120=180, we get the answer as D.

ThatDudeKnows avigutman why do we need them to be congruent? What if it is smaller?

Also how to go about doing it with the equations if we dn't know derivative? We have a quadratic basically with one variable- how do we minimize it?

I'll preface by saying I doubt you can find me an official geometry question that is this tough, so it's probably not worth worrying about unless you are looking to solidify yourself at Q51, which should only be happening if you're already at V46 or better (if not, your time is better spent on V or on other Q topics).

Let's make the base of the rectangle B and the height of the rectangle H.
\(\frac{12}{B}=\frac{H}{5}\), so BH=60. We therefore know that we need two things that multiply to 60

I wish I had a better answer for you, but the minimization element of this feels intuitive to me...we can prove it with partial derivatives, but that's obviously outside the scope of the GMAT...absent that intuition or partial derivatives, we can still figure it out with a little plugging in numbers to see what happens.

Just to start wrapping our head around it, what happens if we make B=12? That would mean H=5. That would mean the big triangle would be 0.5*17*17 = 0.5*289.

What happens if we make B larger? How about B=24? That would mean H=2.5. That would mean the big triangle would be 0.5*29*14.5. Ohhh, that's definitely bigger than when B=12. Doesn't look like we should increase the size of B. Let's try going smaller than 12.

What happens if we make B a little smaller? How about B=10? That would mean H=6. That would mean the big triangle would be 0.5*15*18 = 0.5*270. Heyyyy, that's better than B=12. Let's keep going!

What happens if we make B even smaller? How about B=6? That would mean H=10. That would mean the big triangle would be 0.5*11*22 = 0.5*242. Hey, that's even better! Can we beat that?

What happens if we make B even smaller? How about B=5? That would mean H=12. That would mean the big triangle would be 0.5*10*24 = 0.5*240. Hey, that's even better! But it's a pretty small change from B=6...I wonder if we are either at the optimal point, getting really close, or just passed it over. Either way, we're really close and the answer choices are pretty spread out, so we can probably roll with this even if it's not exactly right (it turns out that it is exactly right).

Even if you'd decided to stop at B=6, you'd have ended up with a triangle area 135 and a rectangle area 60. That's a total of 195. E is absolutely out and you'd have to feel pretty good that you might not have the exact answer but you should be pretty close. So you'd go with D, anyway.

ThatDudeKnows what’s the logic used as to why two triangles have to be similar to minimize ??



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ThatDudeKnows pls clarify why this is the case? What’s the logic or proof for why they need to be congruent




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Since the two triangles are at least similar, we have

12/h=l/5, so

hl=60=area of the rectangle

Since the problem asks for sum of area of larger triangle plus the rectangle, the rectangle area is included twice, meaning the answer is greater than 120, eliminating answers A-C, leaving only D or E.

The area of the upper triangle is

6l and the area of the bottom is 2.5h. Since hl=60, the area of the bottom triangle is also 150/l.

Testing answer D, the sum would be

6l + 150/l + 120 = 180 or

6l^2-60l+150=0 or

L^2-10l+25=0

This is equal to (l-5)^2=0 so

L=5, which means h=12

Testing E isn't necessary since even if it works it is greater than D and so not the minimum

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Kinshook - Can you please explain the relevance of the highlighted portion. I stopped at x=5 and substituted the value to get the answer?

Sonia2023

For minima & maxima both
First derivative should be 0.

For minima
Second derivative should be +ve

For maxima
Second derivative should be -ve.

Please go through basics of calculus

Please read the article below: -
https://byjus.com/jee/maxima-and-minima-in-calculus/

The problem can be solved without using calculus too.

Kinshook - Thank you for sharing the article. I think I got it. Is the below understanding correct?

So when we take out first derivative, we are essentially just coming up with a turning point - that point can be a maxima, minima or just a reflection point. To be sure that the turning point is minima, maxima or just a reflection, we need to use second derivative?

So if second derivative is greater than zero, it confirms that the point is minima. Right?