Question 1182599
The following equation models the rate at which a worker produces so many units after x hours when the worker starts at 8AM (Assume 8AM means x=0)

{{{f(x)=-2x^2+8x+50}}}

At what time is the worker working most efficiently and how many units is the worker producing at this time?

{{{f(x)=-2x^2+8x+50}}}-> this is downward parabola and max is at vertex

so, write equation in vertex form

{{{f(x)=(-2x^2+8x)+50}}}...........factor out {{{-2}}}

{{{f(x)=-2(x^2-4x)+50}}}.......complete square

{{{f(x)=-2(x^2-4x+b^2)-(-2)b^2+50}}}.......{{{b=4/2=2}}}

{{{f(x)=-2(x^2-4x+2^2)+2*2^2+50}}}

{{{f(x)=-2(x-2)^2+8+50}}}

{{{f(x)=-2(x-2)^2+58}}}

=>{{{h=2}}}, {{{k=58}}}=> vertex is at ({{{2}}},{{{58}}})

Assuming {{{8AM}}} means {{{x=0}}}, {{{x=2}}}  will be {{{10AM}}}


the worker working most efficiently at {{{10AM}}}   and producing {{{58}}} units at this time 


{{{ drawing( 600, 600, -10,10, -10, 65, circle(2,58,.15),locate(2,58,v(2,58)),
graph( 600, 600, -10,10, -10, 65, -2(x-2)^2+58)) }}}