Question 1196071
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The referenced URL tells you how you can answer the questions; did you try that?<br>
EF is the same length as DE, so EF=x.<br>
DC is 10, and DE+EC=DC, so EC=10-x.<br>
Then use the Pythagorean Theorem to find an expression for FC:<br>
FC = {{{sqrt((x^2)-(10-x)^2)=sqrt(x^2-(x^2-20x+100))=sqrt(20x-100)}}}<br>
And the area of EFC is one-half base times height:<br>
{{{((10-x)(sqrt(20x-100)))/2}}}<br>
To find the value of x that maximizes the area of triangle EFC, by far the easiest thing to do is graph the area function on a graphing calculator and use the calculator's features to find the answer.<br>
Doing that shows the maximum area to be when x=20/3.<br>
Or you could use calculus, if you know the subject....<br>
(The factor of 1/2 in the area formula is not important for finding the maximum area, so I will ignore it in my work.)<br>
{{{y=(10-x)((20x-100)^(1/2))}}}<br>
Use the product rule and chain rule to find the derivative.<br>
{{{dy/dx=(-1((20x-100)^(1/2)))+(10-x)((1/2)((20x-100)^(-1/2)))(20)}}}
{{{dy/dx=(-(20x-100)^(1/2))+(10(10-x)/(20x-100)^(1/2))}}}<br>
Find where the derivative is equal to zero.<br>
{{{(-(20x-100)^(1/2))+(10(10-x)/(20x-100)^(1/2))=0}}}<br>
{{{((20x-100)^(1/2))=(10(10-x)/(20x-100)^(1/2))}}}<br>
{{{(20x-100)=10(10-x)}}}
{{{20x-100=100-10x}}}
{{{30x=200}}}
{{{x=200/30=20/3}}}<br>
The calculus confirms the answer found using a graphing calculator.<br>