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

Suppose the green rectangle have
Length = L
Breadth = B

Using gthe property of similar triangles, we get
12/L = B/5
LB = 60
Thus, Area of Rectangle = 60 cm^2

Now, Area of Upper triangle = 1/2 * 12 * L
Area of Lower triangle = 1/2 * B * 5

Minimum Area of COMPLETE TRIANGLE = Upper triangle + Lower Triangle + Rectangle

Upper triangle + Lower Triangle = (1/2 * 12 * L) + (1/2 * B * 5)
= 1/2 * (12 L + 5 B)

Using AM >= GM, we get
1/2 * (12 L + 5 B) >= Sq Rt (12 L * 5 B)
1/2 * (12 L + 5 B) >= Sq Rt (60 L * B)
1/2 * (12 L + 5 B) >= Sq Rt (60 * 60) (since LB = 60 as deduced above)
1/2 * (12 L + 5 B) >= 60

Thus , MIn value for Area of COMPLETE TRIANGLE = (Upper triangle + Lower Triangle) + (Rectangle)
= (1/2 * (12 L + 5 B) ) + (LB)
= 60 + 60
= 120

So, Area of RECTANGLE + MIN AREA of Complete Triangle = 60 + 120
= 180 cm^2

(D) is the correct answer


Hope this helps..

In order to minimize the size of the big triangle, we want the two white triangles to be congruent. That means a couple 5-12-13 triangles and a 5x12 rectangle.
Area of the big triangle = 0.5*10*24 = 120
Area of the rectangle = 5*12 = 60
Sum of those two = 180

Answer choice D.

Answer: D

Let the base and height of green rectangle be b and h respectively.

Considering, co-ordinate of right angle of the triangle to be (0,0), the rectangle touches the hypotenuse of triangle at (b,h)

Area of Triangle, A = (1/2)*(12+h)*(b+5) ...Eq 1
We need to optimize the area of triangle.

Slope of hypotenuse = m = (h-0)/(b-5-b) = -h/5
=> h = -5m ...Eq 2
Similarly, m = (12+h-h)/(0-b) = -12/b
=> b = -12/m ...Eq 3

Substituting, value of h and b in Eq 1, we get
A = (1/2)*(12-5m)*(-12/m + 5)
= (1/2)*(-144/m + 60 + 60 - 25m)
= -72/m + 60 - 25m/2

Taking derivative of A, we get
A' = 72/m^2 - 25/2
When A' = 0
=> 72/m^2 = 25/2
=> m^2 = 72*2/25
=> m = -12/5 ... since slope is -ve

Substituting value of m in Eq 2 and Eq 3, we get
h = 12 and b = 5

Therefore, minimum possible area of the triangle = (1/2)*(5+5)*(12+12)
= (1/2)*10*24 = 120

Area of green rectangle = b*h = 5*12 = 60

value of (the minimum possible area of the triangle) PLUS (the area of the green rectangle)
= 120 + 60
= 180

The answer to the question is 180 or choice D. I explained in more detail below.

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.