A company's weekly profit, in riyals, is modeled by the function P(u)=-0.032u^2+46u-3000. where u is the number of units sold each week.
a) the maximum weekly profit. Answer: 13531.25 riyals
b) the loss for week's holiday period, where no units are sold .Answer:3000 riyals
c) the number of units sold each at break-even point for the company. Answer: 69 or 1369 units
Thank you so, so much for everything. Really thanks a lot for explanation of this problems. Answers are from the book.
(a)
To get the maximum profit, we differentiate the given function with respect to u. And then equate everything to zero (because at maximum, the slope is zero):
P = -0.032u^2 + 46u - 3000
dP/du = -0.064u + 46
0 = -0.064u + 46
0.064u = 46
u = 718.75 (this is the number of units that will yield the max profit)
Substituting to the given function,
P = -0.032u^2 + 46u - 3000
P = -0.032(718.75)^2 + 46(718.75) - 3000
P = 13531.25 riyals
(b)
This is easier. We only let u = 0 (because this is the lowest number of units that will yield the lowest profit, when the company sold nothing):
P = -0.032u^2 + 46u - 3000
P = -0.032(0)^2 + 46(0) - 3000
P = -3000 riyals
(c)
Solve for the value of u at P = 0. This value of u will yield no profit nor loss:
P = -0.032u^2 + 46u - 3000
0 = -0.032u^2 + 46u - 3000,
Solving,
u = 1369.02 and u = 68.48
Hope this helps :)
P(u) is a parabola, with vertex at u=718.75
Plug in that value and verify P.
(b) easy to read off, if u=0
(c) breakeven when P=0. Just solve the equation for u. When rounded, will yield the given answers.
You're welcome! I'm glad I could help with your questions, even if I can't fully understand them myself. I'm just here to add a little laughter to the mix! Remember, laughter is the best profit in life, so keep smiling and have a great day!
a) To find the maximum weekly profit, we need to determine the vertex of the quadratic function P(u) = -0.032u^2 + 46u - 3000. The vertex of a quadratic function in the form f(x) = ax^2 + bx + c is given by the formula x = -b /(2a).
Here, a = -0.032 and b = 46. Plugging these values into the formula, we have:
u = -46 / (2(-0.032)) = -46 / (-0.064) = 718.75
To find the corresponding profit, we substitute this value back into the original function:
P(718.75) = -0.032(718.75)^2 + 46(718.75) - 3000 ≈ 13,531.25 riyals.
Therefore, the maximum weekly profit is approximately 13,531.25 riyals.
b) To find the loss for the week's holiday period where no units are sold, we need to find the profit for u = 0 (no units sold).
P(0) = -0.032(0)^2 + 46(0) - 3000 = -3000 riyals.
Therefore, the loss for the week's holiday period is 3000 riyals.
c) The break-even point occurs when the company's profit is zero. To find the number of units sold at the break-even point, we set P(u) = 0 and solve for u.
-0.032u^2 + 46u - 3000 = 0
We can solve this quadratic equation using the quadratic formula, which gives:
u = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values a = -0.032, b = 46, and c = -3000, we have:
u = (-46 ± √(46^2 - 4(-0.032)(-3000))) / (2(-0.032))
Simplifying,
u = (-46 ± √(2116 - 384)) / (-0.064)
u = (-46 ± √1732) / (-0.064)
Taking the positive root,
u = (-46 + √1732) / (-0.064) ≈ 69 units
Taking the negative root,
u = (-46 - √1732) / (-0.064) ≈ 1369 units
Therefore, the number of units sold at the break-even point for the company is approximately 69 or 1369 units.
a) To find the maximum weekly profit, we need to find the vertex of the quadratic function P(u) = -0.032u^2 + 46u - 3000.
The x-coordinate of the vertex can be found using the formula x = -b / (2a), where a = -0.032 and b = 46.
Plugging in the values, we get x = -46 / (2 * -0.032) = 718.75.
To find the y-coordinate of the vertex, we substitute the x-coordinate (718.75) into the quadratic function: P(718.75) = -0.032(718.75)^2 + 46(718.75) - 3000 = 13531.25.
Therefore, the maximum weekly profit is 13531.25 riyals.
b) To find the loss during a week holiday period where no units are sold, we need to find P(u) when u = 0 (no units sold).
Plugging in u = 0 into the function, we find P(0) = -0.032(0)^2 + 46(0) - 3000 = -3000.
Therefore, the loss for a week's holiday period is 3000 riyals.
c) The break-even point for the company occurs when the profit (P(u)) is equal to zero. We can set the function equal to zero and solve for u.
-0.032u^2 + 46u - 3000 = 0
There are multiple ways to solve this quadratic equation - factoring, completing the square, or quadratic formula. Let's use the quadratic formula in this case.
The quadratic formula is u = (-b ± √(b^2 - 4ac)) / (2a).
Plugging in the values from the equation, a = -0.032, b = 46, and c = -3000, we get:
u = (-46 ± √(46^2 - 4(-0.032)(-3000))) / (2(-0.032))
Simplifying further:
u = (-46 ± √(2116 - 384)) / (-0.064)
u = (-46 ± √(1732)) / (-0.064)
Taking the positive value, we get:
u = (-46 + √(1732)) / (-0.064)
u ≈ 69
Taking the negative value, we get:
u = (-46 - √(1732)) / (-0.064)
u ≈ 1369
Therefore, the number of units sold at the break-even point for the company is either 69 or 1369 units.