The supply function for manufacturing a certain item is p(x)=x2+46x−66. The demand function is p(x)=56x+30. If x represents the number of items (in hundreds), what is the optimum number of items to be manufactured?
welcome to QC ~ optimum occurs at supply = demand so set x^2 + 46x - 66 = 56x + 30 and solve for x
ok brb
600?
Hm, not quite can you try combining like terms?
x^2 + 46x - 66 = 56x + 30 for example, you can subtract 56x from both sides, what do you get after that?
i did i put all the xs to the right so i got 10x some how
10x
hm, let's try putting all the x terms on the left instead
can you try subtracting 56x from both sides?
oh nvm i subtracted 46
well i still got 10x
ok
x^2 + 46x - 66 - 56x = ?
yes i just combined like terms now it is x^2+10x-66
almost, since it's 46x - 56x it's -10x not positive 10x so right now we have x^2 - 10x - 66 = 30 now try subtracting 30 from both sides
ok
i got x^2+10x-36
hm, not quite, remember the rules for subtracting signs -66 - 30 = -96
ohh ok
now we have x^2 - 10x - 96 = 0 can you try factoring this? as a hint try to find two numbers that multiply to get -96 and add up to get -10
ok ill try
6 and -16 ?
awesome so our eqn becomes (x+6)(x-16) = 0 splitting this into two parts gives us x + 6 = 0 x - 16 = 0 so what are the two values of x?
x=6 x=-16
hm, not quite x + 6 = 0 subtracting 6 from both sides gives us x = -6, not positive 6 same thing with x - 16 = 0, adding 16 to both sides gives us x = 16
ohh
that being said, since x represents the # of items in the 100's, that means x must be positive so we take the positive solution 16 and multiply by 100 to get the sol'n 1600
oh wow that seems pretty easy now that i know what to do thanks alot have a good day
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