Question 1137649



The table shows the daily production level and profit for a business. Use the quadratic function {{{ y=ax^2+bx+c }}}

to determine the number of units that should be produced each day for maximum profit. What is the maximum daily​ profit?

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{{{x}}}​ (Number of Units Produced​ Daily) | {{{30}}}     |  {{{50 }}}    | {{{100}}}
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{{{y}}}​ (Daily Profit)                   | ​${{{5800}}}  |  ${{{7320}}}  | ​${{{4120}}}


{{{ y=ax^2+bx+c }}}........plug in {{{x=30}}}​ and {{{y=5800}}}​
 
{{{ 5800=a*30^2+b*30+c }}}

{{{ 5800=900a+30b+c }}}..........solve for {{{c}}}

{{{ c=5800-900a-30b }}}.......eq.1



{{{ y=ax^2+bx+c }}}........plug in {{{x=50}}}​ and {{{y=7320}}}​
 
{{{ 7320=a*50^2+b*50+c }}}

{{{ 7320=2500a+50b+c }}}..........solve for {{{c}}}

{{{ c=7320-2500a-50b }}}.......eq.2


{{{ y=ax^2+bx+c }}}........plug in {{{x=100}}}​ and {{{y=4120}}}​
 
{{{ 4120=a*100^2+b*100+c }}}

{{{ 4120=10000a+100b+c }}}..........solve for {{{c}}}

{{{ c=4120-10000a-100b }}}.......eq.3


from eq.1 and eq.2 we have

{{{5800-900a-30b =7320-2500a-50b}}}......solve for{{{ b}}}

{{{50b-30b =7320-2500a-5800+900a}}}

{{{20b =1520-1600a}}}

{{{b =76-80a}}}...........eq.1a


from eq.2 and eq.3 we have

{{{7320-2500a-50b=4120-10000a-100b}}} .....solve for {{{b}}}

{{{100b-50b=4120-10000a-7320+2500a}}} 

{{{50b=-7500a-3200}}}

{{{b=-150a-64}}}..........eq2a


from eq.1a and eq.2a we have


{{{76-80a=-150a-64}}}.....solve for{{{ a}}}

{{{150a-80a=-76-64}}}

{{{70a=-140}}}

{{{a=-140/70}}}

{{{a=-2}}}

go to {{{b =76-80a}}}...........eq.1a, substitute {{{a}}}

{{{b =76-80(-2)}}}

{{{b =76+160}}}

{{{b =236}}}


go to {{{ c=4120-10000a-100b }}}.......eq.3, substitute {{{a}}} and {{{b}}}

{{{ c=4120-10000(-2)-100*236 }}}

{{{ c=4120+20000-23600 }}}

{{{ c=4120-3600 }}}

{{{ c=520 }}}


so, your function is:{{{ y=-2x^2+236x+520 }}}

max is at vertex, so write your function in vertex form {{{y=a(x-h)^2+k}}} where {{{h}}} and {{{k}}} are coordinates of the vertex 



{{{ y=-2x^2+236x+520 }}}....using completing square we have

{{{ y=-2(x^2-118x)+520 }}}

{{{ y=-2(x^2-118x+b^2)-(-2b^2)+520 }}}....{{{b=118/2=59}}}

{{{ y=-2(x^2-118x+59^2)+2*59^2+520 }}}

{{{ y=-2(x-59)^2+6962+520 }}}

{{{ y=-2(x-59)^2+7482 }}}

=>{{{h=59}}} and {{{k=7482}}}


Each day {{{59}}} units should be produced to have a maximum daily profit of ​${{{7482}}}.