I HAVE TO SOLVE THE FOLLOWING PROBLEMS FOR AN ALGEBRA ASSIGNMENT DUE SUNDAY. I HAVE FIGURED OUT MOST OF THE ANSWERS BUT AM STUCK ON A FEW. IF ANYONE CAN TELL ME IF MY EXISTING ANSWERS ARE CORRECT AND POINT ME IN THE RIGHT DIRECTION/GIVE ME A HINT ON THE REST, IT WOULD BE GREATLY APPRECIATED! THANKS IN ADVANCE!

1. Suppose that a market research company finds that at a price of p = $20, they would sell x = 42 tiles each month. If they lower the price to p = $10, then more people would purchase the tile, and they can expect to sell x = 52 tiles in a month¡¯s time. Find the equation of the line for the demand equation. Write your answer in the form p = mx + b. (Hint: Write an equation using two points in the form (x,p)).

HERE'S MY ANSWER...
p=-x+62
IS IT CORRECT?

2. A company¡¯s revenue is the amount of money that comes in from sales, before business costs are subtracted. For a single product, you can find the revenue by multiplying the quantity of the product sold, x, by the demand equation, p.

Substitute the result you found from part a into the equation R = xp to find the revenue equation. Provide your answer in simplified form.

HERE IS MY ANSWER...
R=xp
R=x(-x+62)=-x^2+62x
IS IT CORRECT?

3. The costs of doing business for a company can be found by adding fixed costs, such as rent, insurance, and wages, and variable costs, which are the costs to purchase the product you are selling. The portion of the company¡¯s fixed costs allotted to this product is $300, and the supplier¡¯s cost for a set of tile is $6 each. Let x represent the number of tile sets.

3A. If b represents a fixed cost, what value would represent b?

HERE IS MY ANSWER...
b=300
IS IT CORRECT?

3B. Find the cost equation for the tile. Write your answer in the form C = mx + b.

HERE IS MY ANSWER...
C=mx+b
C=6x+300
IS IT CORRECT?

3C. The profit made from the sale of tiles is found by subtracting the costs from the revenue. Find the Profit Equation by substituting your equations for R and C in the equation . Simplify the equation.

HERE'S MY ANSWER...
P=56x-x^2-300
IS IT CORRECT?

3D. What is the profit made from selling 20 tile sets per month?

HERE'S MY ANSWER...
P=56x-x^2-300
P=(56∙20)-20^2-300
P=1120-400=720-300=420
IS IT CORRECT?

3E. What is the profit made from selling 25 tile sets each month?

HERE'S MY ANSWER...
P=56x-x^2-300
P=(56∙25)-25^2-300
P=1400-625=775-300=475
IS IT CORRECT?

3F. What is the profit made from selling no tile sets each month? Interpret your answer.

HERE'S MY ANSWER...
P=56x-x^2-300
P=(56∙0)-0^2-300
P=0-0-300=-300
IS IT CORRECT?

3G. Use trial and error to find the quantity of tile sets per month that yields the highest profit.

I'M CONFUSED ON THIS ONE BUT HERE'S MY ANSWER ANYHOW...
6x¡Ü300
6x/6¡Ü300/6
x¡Ü50
P=-50^2+(56∙50)-300
P=-2500+2800=300-300=0
IS IT CORRECT?

3H. How much profit would you earn from the number you found in part 3F?

MY ANSWER...
You wouldn¡¯t earn a profit (????)
IS IT CORRECT?

3I. What price would you sell the tile sets at to realize this profit (hint, use the demand equation)?

HERE'S MY ANSWER...
p=-x+62
-300=-x+62
x=362
IS IT CORRECT?

4. The break even values for a profit model are the values for which you earn $0 in profit. Use the equation you created in question one to solve P = 0, and find your break even values.

HERE IS MY ANSWER...
P=-x+62
0=-x+62
x=62 (????)

IS IT CORRECT?

1.

p=-x+62
is correct

2. -x^2+62 x
is correct

3a.
b = 300
is correct

3b.
c = 6x+300
is correct

3c.
P=56x-x^2-300
is correct but usually written
P = -x^2 + 56 x - 300

3d.
P = -400 + 1120 - 300
P = $520
correct

3e.
P = -625 + 1400 - 300
P = 475
correct

3f.
-300 is correct
You still had to pay the rent, etc. even though selling nothing.

3g.
I will cheat and use calculus
dP/dx = 0 at maximum
-2x+56 = 0
x = 28 at vertex of parabola
Now the way without calculus
P = -x^2 + 56 x - 300 is a parabola
I can find the axis of symmetry by completing the square
- x^2 + 56 x = (P+300)
x^2 - 56 x = -(P+300)
x^2 -56 x + 784 = -P + 484
(x-28)^2 = -P +484
so
Maxi at x = 28 checks :)

3h. THEY MEAN IN PART 3 G ! I am sure.
We can see from the parabola that P = 484 when x = 28 but check
P =-28^2 + 56(28) - 300
P = -784 + 1526 - 300
P = $484 sure enough

3i.
p = -x+62
=-28+62
= $34 per tile set price

4. You have confused little p, price , with big P, profit
P = -x^2 + 56 x - 300
when is P = 0?
solve quadratic
x^2 - 56 x + 300 = 0
(x-2)(5-50) = 0
x = 2 and x = 50 are break even

Thank you so much for your help! I see where I got confused now. Thanks :o)

4. You have confused little p, price , with big P, profit

P = -x^2 + 56 x - 300
when is P = 0?
solve quadratic
x^2 - 56 x + 300 = 0
(x-6)(x-50) = 0
x = 6 and x = 50 are break even

Where did the 56 come from in part 3c

I'm confused where does P=56x-x^2-300

where does the 56x come from??? shouldn't it be P= x^2+62-300

How did you do on this assignment. Did you get these right, because I am confused about the part where it asks you to plug in P=R-C. Where R=-x^2+62 and C=6x+300. However, you only put P=(-x^2+62)+300. What happened to the 6x in C?

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This is cheating to get answers to graded work online

1. For the demand equation, you correctly used the two given points (42, 20) and (52, 10) to find the equation of the line. However, your answer is incorrect. The equation of the demand equation should be p = -x + 62.

2. For the revenue equation, you correctly multiplied the quantity sold (x) by the demand equation (p). Your answer of R = -x^2 + 62x is correct.

3A. For the fixed cost, you correctly identified that b = 300.

3B. For the cost equation, you correctly used C = mx + b, and your answer of C = 6x + 300 is correct.

3C. For the profit equation, you correctly subtracted the cost equation from the revenue equation. Your answer of P = -x^2 + 56x - 300 is correct, not P = 56x - x^2 - 300.

3D. Your calculation for the profit made from selling 20 tile sets per month is correct. The profit would be 420.

3E. Your calculation for the profit made from selling 25 tile sets per month is correct. The profit would be 475.

3F. Your calculation for the profit made from selling no tile sets per month is correct. The profit would be -300. This means that you would incur a loss of $300.

3G. For finding the quantity of tile sets per month that yields the highest profit, you need to calculate the vertex of the parabola represented by the profit equation. The x-coordinate of the vertex is given by -b/2a, where a and b are the coefficients of the quadratic equation. In this case, a = -1 and b = 56. So, the x-coordinate of the vertex is -56/(2*-1) = 28. Plug this value back into the profit equation to find the profit at this quantity.

3H. The profit earned from the quantity found in part 3F would be -300.

3I. To find the price at which you would sell the tile sets to realize a profit of -300, plug the quantity (x = 0) into the demand equation p = -x + 62. The price would be 62.

4. For the break-even values, you need to solve the profit equation P = 0. You correctly set P = -x + 62 and solved for x. Your answer x = 62 is correct. This means that the break-even point is when you sell 62 tile sets per month.