Wow, this was one of my first posts a loooong time ago. I believe I was doing my best to try to absorb a million different concepts.
Looking at the problem now, I believe it is solved much more easily using a concept that can be applied to certain optimization- type problems.
Concept:
Given 2 positive values, it will always be true that —
(arithmetic mean) >/= (geometric mean)
Here is a link to read up on the concept:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2015/0 ... at-part-v/Offhand, I can think of one or two official problems with which you could make use of this concept.
The other concept that is applicable in this problem is the use of the Intercept Form for the Graph of a Straight Line in the coordinate plane.
The graph of the straight line can be expressed in the form:
X / (X - intercept) + Y / (Y-intercept) = 1
1st) we can call the X- intercept point (p , 0) and the Y-Intercept (0 , q)
The equation for the graph of this line in Intercept Form would be:
(X / p) + (Y / q) = 1
Furthermore, since the line passes through the point (3 , 48), that coordinate point will satisfy the equation
Substituting (3 , 48) we get —
(3 / p) + (48 / q) = 1 ——(equation 1)
(2nd). Since p and q are positive values, we can apply the (A.M.) >/= (G.M.) concept
[(3/p) + (48/q)] / 2 >/= sqrt( (3/p) (48/q) )
Substitute the value 1 from equation 1 above in the numerator of the arithmetic mean (A.M.) on the left side of the inequality we get:
1 / 2 >/= sqrt( 144 / PQ )
Then square both sides
1 / 4 >/= 144 / PQ
PQ >/= (4) (144)
PQ >/= 576
So the product of the X intercept and the Y intercept must be greater than or equal to 576
Minimum value = 576
I wish I could help you with what I wrote almost 2 years ago. However, I sometimes forget what I had for breakfast 😂
Thank you for the kind words!
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