Question

If three sides of a trapezoid are each 6 inches long, how long must the fourth side be if the area is a maximum

Answer 1

is this calculus ?

Answer 2

we know the top and base can't both be 6 , because then we would not have a trapezoid... the top and base of a trapezoid have different lengths.

so we have

Answer 3

12

Answer 4

area as solved above A=(6+x)h=(6+x)sqrt (36-x^2)
\[Let s=A ^{2}=\left( 6+x \right)\left( 36-x ^{2} \right)\]
\[\frac{ ds }{dx }=\left( 6+x \right)^{2}\left( -2x \right)+\left( 36-x ^{2} \right)*2\left( 6+x \right)\]
\[=2\left( 6+x \right)[-x \left( 6+x \right)+(6+x)(6-x)]\]
\[=2\left( 6+x \right)^{2}\left[ -x+6-x \right] \]
\[=-4\left( 6+x \right)^{2}\left( x-3 \right) \]
\[put \frac{ ds }{dx }=0 and find stationary values ,when \frac{ d ^{2}s }{ dx ^{2} } is <0\]
for any value then that value is for maxima.
solve it ,if any problem you can ask me.