Question 342259
With
{{{x = 1700 -100p}}}
and
{{{7000 = xp}}}
we have a system of two equations with two variables. Since the second equation is of degree 2 (because of the x*p term), we will use the Substitution Method. (With second degree equations other methods are often either not possible or not practical.)<br>
Since we already have the first equation solved for x, we will use that and substitute into the second equation:
{{{7000 = (1700-100p)p}}}
To solve this we will start by simplifying:
{{{7000 = 1700p-100p^2)}}
Since this is a quadratic equation we want one side to be zero:
{{{100p^2 -1700p + 7000 = 0}}}
We can simplify the equation by dividing both sides by 100:
{{{p^2 -17p + 70 = 0}}}
Now we can factor (or use the Quadratic Formula):
{{{(p-7)(p-10) = 0}}}
From the Zero Product property we know that:
{{{p-7 = 0}}} or {{{p-10 = 0}}}
solving these we get:
p = 7 or p = 10
Now we find the x for each of these p's. We'll use the first equation and substitute in the value for p:
For p = 7:
{{{x = 1700 - 100(7) = 1700 - 700 = 1000}}}
For p = 10:
{{{x = 1700 - 100(10) = 1700 - 1000 = 700}}}<br>
So we have two solutions which generate $7000 revenue:
When the price is $7, 1000 calculators will be sold.
When the price is $10, 700 calculators will be sold.<br>
If it is cheaper to build 700 calculators than it would be to build 1000 calculators, the greatest profit would come from making 700 calculators and selling them for $10 each.