Question

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

Answer 1

to find the value of 'h', use pythagorean theorem:
Isolating triangle at the left:

\[x ^{2} + h ^{2} = 6^{2}\]
\[h = \sqrt{6^{2}-x ^{2}}\]
NOTE that the area of a triangle is designated by the formula:
\[A = \frac{ 1 }{ 2 } bh\] or in your case
\[A = \frac{ 1 }{ 2 } xh\]
but since there two triangles then we may have the area twice or
\[A =\frac{ 1 }{ 2 } xh (2)\]
\[A = xh\]
Isolating the rectangle inside you'll have:

Area for triangle is:
\[A = bh\] or in that case
\[A = 6h\]

Answer 2

So to sum it up, you'll have the total area as:
Area of the two triangles + area of the rectangle
\[xh + 6h\]
factoring out h:
\[h(x+6)\]
having the value for 'h' above as
\[h = \sqrt{6^{2}-x ^{2}}\]
you may substitute to find the total area
\[A _{T} = (\sqrt{6^{2}-x ^{2}})(x+6)\]
\[A _{T} = \sqrt{6^{2}(x ^{2}+12x+36) - x ^{2}(x ^{2}+12x+36)}\]

Answer 3

\[A _{T} = \sqrt{36x ^{2}+432x + 1296 -x ^{4}-12x ^{3}-36x ^{2}}\]
combining like terms:
\[A _{T} = \sqrt{-x ^{4}-12x ^{3} +432x +1296}\]
You are now ready to differentiate.
Differentiate both sides

Answer 4

differentiating you'll get:
\[\frac{ dA }{ dx } = \frac{ 1 }{ 2 }(-x ^{4}-12x ^{3}+432x+1296)^{-\frac{ 1 }{ 2 }}(-4x ^{3}-36x ^{2}+432)\frac{ d }{ dx }\]

Answer 5

@robtobey will you be kind enough to recheck the above computations if in any case there's an error

Answer 6

\[\frac{ dA }{ dx } = \frac{ 1 }{ 2 } \frac{ (−4x ^{3}−36x ^{2}+432) }{ (−x ^{4}−12x ^{3}+432x+1296)^{\frac{ 1 }{ 2 }} }\frac{ d }{ dx }\]

Answer 7

equating dA/dx to zero and cross multiplying the denominator:
\[0 = (-4x ^{3}-36x ^{2}+432)\frac{ d }{ dx } \]

Answer 8

or you'll have it as:
\[4x ^{3}+36x ^{2}=432\]

Answer 9

getting the value of x:
x = 3, -6
take the positive value
x=3
so the length of the other side must be:
\[length = 2x + 6\]
\[length = 2(3) + 6\]
\[length = 12\]