The number x of stereo speakers a retail chain is willing to sell per week at a price of p dollars is given by. Find the supply and instantaneous rate of change of supply when the price is 75 dollars?

The question is not complete and the complete question says:

The number  x  of stereo speakers a retail chain is willing to sell per week at a price of  p  dollars is given by x=80√(p+27) − 410 Find the supply and instantaneous rate of change of supply when the price is  75  dollars

A) Approximately 398 stereo speakers

B) Instantaneous rate of change = 3.961

Step-by-step explanation:

From the question, the number of speakers is given as;

x=80√(p+27) − 410

A) Thus at a price of $75,

The amount supplied will be;

x = 80√(75+27) − 410

x = 80√(102) - 410

x = (80 x 10.0995) - 410

x = 807.96 - 410 = 397.96 ≈ 398 stereo speakers

B) The instantaneous rate of change is given as;

dx/dp of the supply equation.

Thus, let's rearrange the supply equation first before differentiating.

So, x = 80√(p+27) − 410

x = 80[(p +27)^(1/2)] - 410

So, dx/dp = [(80 x 1/2) (p + 27)^(-1/2)]

dx/dp = 40/√(p+27)

So, at price of $75,

dx/dp = 40/√(75+27) = 40/√102

= 40/10.0995 = 3.961

The answer is d, (20 - 10, 10) ∪ (30, 20 + 10)

Step-by-step explanation:

We know that we have 80 feet of fencing available, and that we really only need to have fencing around 3 sides of the garden.

80 feet will equal the sum of the horizontal side length (long one) and the two vertical side lengths. Represent this as 80 = 2x + y, where x is the short length, and y is the length of the long side of fencing. There is a 2x because we need two short sides of fencing and only one long side (y) because we don't need another long side of fencing closest to our house.

We are looking for possible lengths of the long side, y. Manipulate the equation such that y is isolated:

y = -2x + 80

Now the tricky part. We know that the area of the garden must be greater than 400 and less than 600. We can represent this situation with:

400 < xy < 600, where xy is the area (x represents width/short sides, and y represents length/the long side, remember that area = length times width)

This compound inequality is impossible to solve. But, we know that we have already written y in terms of x from before. (y = -2x + 80) Using this, we can substitute y in this compound inequality with this equation. So,

400 < x(-2x + 80) < 600

= 400 < -2x² + 80x < 600

Now, we go about the regular process in solving quadratic inequalities, in this case, we have a compound inequality.

Solving for the left side, we get 0 < -2x² + 80x - 400 (I subtracted 400 from both sides) Keep this inequality in mind.

Solving for the right side, we get -2x² + 80x - 600 < 0, which is the same as 0 > -2x² + 80x - 600. (I subtracted 600 from both sides) Keep this inequality in mind.

You can get crazy with the quadratic equation, but I didn't do that on this quiz, as I might not have even finished on time lol. I just chose to graph both quadratic functions on the Edgenuity graphing calculator, and looked at where each of the two functions had x values that made the inequality true:

where the first function, -2x² + 80x - 400, had positive y values. This starts after x = 5.858 (equivalent to 20 - 10 from the answer choices lol)

a little bit to the right of that, we see where the second function starts to have positive values, which is after x = 10. but remember, we don't care about when this function has positive y values, we care about when it has negative y values.

this forms the first interval, (20 - 10, 10). 10 is there because up until x=10, the inequality is satisfied.

to the right of this, we see that the second function (looks a little more squished) touches the x axis again at 30, and after 30, the function becomes true according to the inequality and starts to have negative values, and will continue to so for positive infinity.  Keep 30 in mind.

and a little to the right of that, we have that the first function (the wider one) touches the x axis at x = 34.142, equivalent to 20 + 10 from the answer choices(I punched this into a calculator to see) and then begins to have negative values after this, making the inequality not true.

This forms the second interval, (30, 20 + 10). That's how I got my answer.

Think back to the Edgenuity lesson on function modeling. We had to find areas on a graph where all functions were satisfied called the feasible region. This problem builds on those skills, and the "feasible regions" if you're looking at the two functions graphed, are the two areas where there is a thin gap between the line of the first and second function. The answer to this problem is the "domains" if you'd like to think of it that way, of these two areas where both inequalities are true.

I hope this helps. Edgenuity precal is getting harder and harder.