# Difference between revisions of "1962 AHSME Problems/Problem 37"

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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> or <math>\boxed{\textbf{(D)}},</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 |

## Latest revision as of 22:51, 24 November 2020

## 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