Were asked to use the methods of this section to answer a question about cost, revenue and profit. We're told that the revenue and cost equations for a product are okay. Are the revenue equals x Times 75 minus 0.5 x and see The cost equation is 30 x plus 250,000 where both are and see are measured in dollars and X represents the number of units sold. Okay were asked how many units must be sold to obtain a profit of at least $750,000 and then were asked, what is the price per unit now? First to find the number of units sold to obtain a profit of at least $750,000 we first want to find an equation for the profit. So I called it the Prophet P is three revenue ar minus the cost. Putting these equations together. This is X Times 75 minus 0.5 X minus 30 X plus 250,000. Yeah. Okay, now we can simplify this equation. This is the same as so we have 75 x minus 30 x We have negative point 000 five X squared plus 45 x minus 250,000. So now we have this equation for the profit. Since our prophet must be at least 750,000, we want to solve the nonlinear inequality. He is greater than or equal to 750,000 Solve this inequality well, notice that this is the same as the inequality. Negative 0.5 X squared plus 45 x minus 250,000 is greater than or equal to 750,000. Now we have a polynomial inequality. To solve this inequality, we want to isolate a polynomial on one side and zero on the other. Then you want to find the zeros of this polynomial. Then we want to test values lying in the intervals between the zeros. And finally, we want to determine whether or not the intervals lie in the solution set so we can rewrite this as negative point 0005 X squared plus 45 x and then we have minus boy of 250,000, minus 750,000 which is minus one million, is greater than or equal to zero. Okay, now this is a quadratic polynomial to find the zeros and use the quadratic inequality. Sorry. Quadratic formula. This is the zeros air going to be negative. 45 plus or minus. The square root of 45 squared minus four times negative 0.0 Five times one million all over two times negative. 20.5 Now simplify this. The zeros are approximately 45.97 and 174 0.3 to two decimal places. Notice that our inequality here is not strict. Therefore, the zeros are also solutions to the inequality. Yeah. Now, with these two zeros, these are the only two zeros of our quadratic polynomial. So we can rewrite it as negative 20.5 times. And then we have the two factors X minus approximately 45.97 and x minus 174.3 And the inequality is still greater than or equal to zero. Of course, we can simplify by dividing both sides by negative 0.5 Because this is less than zero. We have to switch the sign of the inequality. So this becomes x minus 45.97 times X minus 174.3 is now less than or equal to zero. So we have this new form of our inequality Now, going back to our zeros. These two zeros divide the real number line into three parts. So the first part, these air all the numbers less than the smaller zero. 45 0.97 As a test value, consider X equals 45. We plug this in. We have see 45 minus 45.97 times 45 minus 174 0.3 Now it's clear that this first term is going to be less than zero and the second term is also going to be less than zero. Therefore, it follows that the product of these two terms is going to be greater than zero. Therefore, the test value is not a solution. Polynomial is only change signs at their zeros and therefore it follows that none of the points in this interval our solutions second reason to consider are all points between 0 45.97 and the other zero. 174.3 To test value, consider 46. Plug this in. We have 46 minus 45.97 and then 46 minus 174.3 Now it's clear that this first factor is going to be greater than zero. The second factor will be less than zero. So if a product of a positive and a negative and therefore the product is going to be less than zero as well. Therefore, the test value is a solution to our inequality again because this is a polynomial inequality. It follows that all the values in this interval our solutions. Finally, we consider the values greater than 174.3 As a test value. Consider X equals 175. Well, we have 40 Sorry. 175 minus 45.97 times 1 75 minus 1 74 0.3 Now, clearly the first factor is going to be created than zero. And so the second factor and therefore the product is also going to be greater than zero. Therefore, the test values not a solution and by the previous discussion none of the points in the interval are either. So putting this together we hit the solution. Sets of this inequality is going to be well. We have left bracket 45 point 97 since this is a solution and then 1 74.3 right bracket. Since this is also a solution, however, let's recall what X represents X is the number of units sold. Therefore, X must be a natural number. So we have to sell a positive number of units or, of course, zero units. This means that numbers like 45.97 which are fractional, are not solutions. So to remedy this, we simply change our answer so that the end points are both integers which lie inside the interval. So we have that 45.97 is of course closest to the image of 46 but still less than it. And 174 0.3 is closest to, but still greater than 174. And therefore it follows that the number of units must be sold to obtain a profit of at least $750,000 is going to be a number of units between 46 and 174. That's our answer in set notation. Now we're asked to find the price per unit No.