Question

if three sides of trapezoid are each 6 inches long, how long must the fourth side be if the area is maximum? ans. is 12

Answer 1

use the formula for the area of a trapezoid...then to maximize you take the first derivative set it equal to 0, and find critical values...does this sound familiar

Answer 2

what class is this for?

Answer 3

\[A = \frac{1}{2}h(x+6)\]
from the right triangle we can define "h" in terms of "x"
\[h^{2} +\frac{(x-6)^{2}}{4} = 36\]
\[h = \frac{\sqrt{144 - (x-6)^{2}}}{2}\]
sub into Area equation
\[A = \frac{1}{4}(x+6) \sqrt{144-(x-6)^{2}}\]
differentiate and set equal to 0
\[\frac{dA}{dx} = \frac{1}{4}\sqrt{144-(x-6)^{2}} -\frac{x-6}{\sqrt{144-(x-6)^{2}}}=0\]
solve for x
\[144 - (x^{2}-12x+36) = 4x-24\]
\[x^{2} -8x-132 = 0\]
\[x = 4+2\sqrt{37}\]

Answer 4

sorry i made an error with the derivative...obviously the answer must be 12