Question 881046
The quadratic has the form,
{{{q=ap^2+bp+c}}}
{{{1500=a(30)^2+b(30)+c}}}
1.{{{900a+30b+c=1500}}}
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Similarly,
2.{{{1600a+40b+c=3600}}}
and
3.{{{2500a+50b+c=6300}}}
Subtracting eq. 1 from eqs. 2 and 3,
{{{2500a+50b+c-900a-30b-c=6300-1500}}}
{{{1600a+20b=4800}}}
4.{{{80a+b=240}}}
and
{{{1600a+40b+c-900a-30b-c=3600-1500}}}
{{{700a+10b=2100}}}
5.{{{70a+b=210}}}
Substituting from eq. 4 into eq. 5,
{{{70a+(240-80a)=210}}}
{{{-10a+240=210}}}
{{{-10a=-30}}}
{{{a=3}}}
Then,
{{{80(3)+b=240}}}
{{{b=0}}}
and finally,
{{{1600(3)+40(0)+c=3600}}}
{{{c=-1200}}}
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{{{q=3p^2-1200}}}
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{{{0=3p^2-1200}}}
{{{3p^2=1200}}}
{{{p^2=400}}}
{{{p=20}}}
20 is the lowest price of the item (usually it gets lower as quantity increases, in this case, it's more expensive to make more, that's counterintuitive) 
{{{graph(300,300,-10,50,-500,4500,3x^2-1200)}}}