Difference between revisions of "1962 AHSME Problems/Problem 37"
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<math> \textbf{(A)}\ \frac{1}2\qquad\textbf{(B)}\ \frac{9}{16}\qquad\textbf{(C)}\ \frac{19}{32}\qquad\textbf{(D)}\ \frac{5}{8}\qquad\textbf{(E)}\ \frac{2}3 </math> | <math> \textbf{(A)}\ \frac{1}2\qquad\textbf{(B)}\ \frac{9}{16}\qquad\textbf{(C)}\ \frac{19}{32}\qquad\textbf{(D)}\ \frac{5}{8}\qquad\textbf{(E)}\ \frac{2}3 </math> | ||
+ | ==Solution== | ||
− | |||
− | |||
Let <math>AE=AF=x</math> | Let <math>AE=AF=x</math> | ||
<math>[CDFE]=[ABCD]-[AEF]-[EBC]=1-\frac{x^2}{2}-\frac{1-x}{2}</math> | <math>[CDFE]=[ABCD]-[AEF]-[EBC]=1-\frac{x^2}{2}-\frac{1-x}{2}</math> | ||
Line 14: | Line 13: | ||
So <math>[CDFE]\le \frac{5}{8}</math> | So <math>[CDFE]\le \frac{5}{8}</math> | ||
Equality occurs when <math>AE=AF=x=\frac{1}{2}</math> | Equality occurs when <math>AE=AF=x=\frac{1}{2}</math> | ||
− | So maximum value is <math>\frac{5}{8}</math> | + | So the maximum value is <math>\frac{5}{8}</math> |
==Solution 2== | ==Solution 2== | ||
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where the first inequality was from the Trivial Inequality, and the second came from dividing both sides by <math>2</math>, which does not change the inequality sign. | where the first inequality was from the Trivial Inequality, and the second came from dividing both sides by <math>2</math>, which does not change the inequality sign. | ||
− | Thus, the maximum area is <math>\boxed{\frac{5}{8}},</math> when <math>\frac{(x-\frac{1}{2})^2}{2} = 0.</math> | + | Thus, the maximum area is <math>\boxed{\frac{5}{8}}</math> or <math>\boxed{\textbf{(D)}},</math> when <math>\frac{(x-\frac{1}{2})^2}{2} = 0.</math> |
+ | |||
+ | ==Solution 3 (Calculus)== | ||
+ | |||
+ | Let <math>AE = AF = x</math>. | ||
+ | |||
+ | The area of the quadrilateral <math>CDFE</math> can be expressed as <math>A = 1 - \frac{x^2}{2} - \frac{(1-x)}{2}</math>. | ||
+ | |||
+ | By taking the derivative of <math>A</math>, we get <math>A' = -x + \frac{1}{2}</math>. | ||
+ | |||
+ | We make <math>A' = 0</math> and get the critical point <math>x = \frac{1}{2}</math>. | ||
+ | |||
+ | Substituting <math>x = \frac{1}{2}</math> , the maximum area is <math>\boxed{\frac{5}{8}}</math> or <math>\boxed{\textbf{(D)}}</math>. | ||
+ | |||
+ | ~belanotbella | ||
+ | |||
+ | ==Solution 4== | ||
+ | Plot the points of the quadrilateral (with the bottom left corner of the square being at the origin) to plug into shoelace formula: | ||
+ | <cmath>(0, 0), (0, 1-x), (x, 1), (1, 0)</cmath> | ||
+ | So, | ||
+ | <cmath>\frac{-x^2+x+1}{2}</cmath> | ||
+ | Trying <math>1/2</math> first, you get <math>5/8</math>. Any larger or smaller value gives an answer less than <math>5/8</math>. Therefore, the answer is <math>\boxed{D}</math>. | ||
− | ~ | + | ~MC413551 |
Latest revision as of 09:08, 13 April 2023
Problem
is a square with side of unit length. Points
and
are taken respectively on sides
and
so that
and the quadrilateral
has maximum area. In square units this maximum area is:
Solution
Let
Or
As
So
Equality occurs when
So the maximum value is
Solution 2
Let us first draw a unit square.
We will now pick arbitrary points
and
on
and
respectively. We shall say that
Thus, our problem has been simplified to maximizing the area of the blue quadrilateral.
If we drop an altitude from
to
, and call the foot of the altitude
, we can find the area of
by noting that
.
We can now finish the problem.
Since and
, we have:
To maximize this, we compete the square in the numerator to have:
Finally, we see that , as:
So,
where the first inequality was from the Trivial Inequality, and the second came from dividing both sides by , which does not change the inequality sign.
Thus, the maximum area is
or
when
Solution 3 (Calculus)
Let .
The area of the quadrilateral can be expressed as
.
By taking the derivative of , we get
.
We make and get the critical point
.
Substituting , the maximum area is
or
.
~belanotbella
Solution 4
Plot the points of the quadrilateral (with the bottom left corner of the square being at the origin) to plug into shoelace formula:
So,
Trying
first, you get
. Any larger or smaller value gives an answer less than
. Therefore, the answer is
.
~MC413551