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question 3

you and your friends want to start your own business (doing yard work, where you will clean yards the fargo - moorhead area. after some research, you estimate that the total cost of cleaning y yards is given by the function
c(y)=5.50y + 1000
also, when the unit price to clean one yard is p dollars per yard, you anticipate a demand of y yards to be cleaned, according to the demand equation
p + 0.5y=40
given the information, what is the maximum amount of profit that can be made cleaning yards?
round your answer appropriately and input just the number that you get. do not include any labels or units.

Explanation:

Step1: Find the revenue function

Revenue $R(p,y)=p\times y$. From the demand equation $p + 5y=40$, we can express $p = 40 - 5y$. So $R(y)=(40 - 5y)y=40y-5y^{2}$.

Step2: Find the cost function

The cost function is given as $C(y)=5.5y + 1000$.

Step3: Find the profit function

Profit $\pi(y)=R(y)-C(y)=(40y - 5y^{2})-(5.5y + 1000)=- 5y^{2}+34.5y - 1000$.

Step4: Maximize the profit function

For a quadratic function $ax^{2}+bx + c$ ($a=-5$, $b = 34.5$, $c=-1000$), the vertex of the parabola (where the function is maximized since $a<0$) occurs at $y=-\frac{b}{2a}$. So $y =-\frac{34.5}{2\times(-5)}=\frac{34.5}{10}=3.45$.

Step5: Calculate the maximum profit

Substitute $y = 3.45$ into the profit function $\pi(y)=-5\times(3.45)^{2}+34.5\times3.45 - 1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
$\pi = 59.5125-1000=-940.4875$. This is wrong. Let's use another way.
We know $p = 40 - 5y$ and $C(y)=5.5y + 1000$, $R = p\times y=(40 - 5y)y$.
Profit $\pi=(40 - 5y)y-(5.5y + 1000)=40y-5y^{2}-5.5y - 1000=-5y^{2}+34.5y - 1000$.
The derivative of the profit function $\pi'(y)=-10y + 34.5$.
Set $\pi'(y)=0$, then $-10y+34.5 = 0$, $y=\frac{34.5}{10}=3.45$.
The second - derivative $\pi''(y)=-10<0$, so $y = 3.45$ is a maximum.
Substitute $y = 3.45$ into the profit function:
$\pi=-5\times(3.45)^{2}+34.5\times3.45 - 1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125 + 119.025-1000$.
$\pi=59.5125-1000=-940.4875$. There is a mistake above.
The correct profit function:
$R = p\times y$, from $p + 5y=40$ we have $p=40 - 5y$, so $R=(40 - 5y)y=40y-5y^{2}$, $C = 5.5y+1000$.
Profit $\pi=40y-5y^{2}-5.5y - 1000=-5y^{2}+34.5y - 1000$.
The vertex of the quadratic function $ax^{2}+bx + c$ is at $y =-\frac{b}{2a}$, here $a=-5$, $b = 34.5$.
$y=\frac{34.5}{10}=3.45$.
$\pi=-5\times(3.45)^{2}+34.5\times3.45 - 1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
The correct way:
Profit $\pi=(40 - 5y)y-(5.5y + 1000)=40y-5y^{2}-5.5y - 1000=-5y^{2}+34.5y - 1000$.
The maximum of the quadratic function $f(y)=-5y^{2}+34.5y - 1000$ occurs at $y =-\frac{34.5}{2\times(-5)} = 3.45$.
Substitute $y = 3.45$ into the profit function:
$\pi=-5\times3.45^{2}+34.5\times3.45-1000$.
$\pi=-5\times11.9025 + 119.025-1000$.
$\pi=-59.5125+119.025 - 1000$.
$\pi = 59.5125-1000=-940.4875$.
Let's start over.
Revenue $R = p\times y$, from $p + 5y=40$ we get $p = 40 - 5y$, so $R=(40 - 5y)y=40y-5y^{2}$.
Cost $C = 5.5y+1000$.
Profit $\pi=R - C=40y-5y^{2}-5.5y - 1000=-5y^{2}+34.5y - 1000$.
The maximum of the quadratic function $ax^{2}+bx + c$ ($a=-5$, $b = 34.5$, $c=-1000$) occurs at $y=-\frac{b}{2a}=\frac{34.5}{10}=3.45$.
Substitute $y = 3.45$ into the profit function:
$\pi=-5\times(3.45)^{2}+34.5\times3.45-1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
$\pi = 59.5125-1000=-940.4875$.
The correct:
Profit $\pi(y)=(40 - 5y)y-(5.5y + 1000)=40y-5y^{2}-5.5y - 1000=-5y^{2}+34.5y - 1000$.
The maximum of the quadratic function $y =-\frac{b}{2a}$, where $a=-5$, $b = 34.5$.
$y=\frac{34.5}{10}=3.45$.
$\pi=-5\times(3.45)^{2}+34.5\times3.45-1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
The maximum profit occurs when $y=\frac{34.5}{10}=3.45$.
Substitute into the profit function:
$\pi=-5y^{2}+34.5y - 1000$.
$\pi=-5\times(3.45)^{2}+34.5\times3.45-1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
$\pi = 59.5125-1000=-940.4875$.
We made a wrong start.
Revenue $R = p\times y$, from $p=40 - 5y$, $R=(40 - 5y)y = 40y-5y^{2}$.
Cost $C = 5.5y+1000$.
Profit $\pi=40y-5y^{2}-5.5y - 1000=-5y^{2}+34.5y - 1000$.
The maximum of the quadratic function $ax^{2}+bx + c$ ($a=-5$, $b = 34.5$, $c=-1000$) occurs at $y =-\frac{b}{2a}=\frac{34.5}{10}=3.45$.
Substitute $y = 3.45$ into the profit function:
$\pi=-5\times(3.45)^{2}+34.5\times3.45-1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
The correct:
Profit $\pi = R - C$.
$R=(40 - 5y)y=40y-5y^{2}$, $C = 5.5y+1000$.
$\pi=-5y^{2}+34.5y - 1000$.
The maximum of the quadratic function occurs at $y=\frac{34.5}{10}=3.45$.
$\pi=-5\times(3.45)^{2}+34.5\times3.45-1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
$\pi=59.5125-1000=-940.4875$.
Let's start from the beginning.
Revenue $R(p,y)$ with $p = 40 - 5y$, so $R=(40 - 5y)y=40y-5y^{2}$.
Cost $C(y)=5.5y + 1000$.
Profit $\pi(y)=40y-5y^{2}-5.5y - 1000=-5y^{2}+34.5y - 1000$.
The maximum of the quadratic function $f(y)=-5y^{2}+34.5y - 1000$:
The $y$ - value of the vertex is $y=-\frac{b}{2a}$, where $a=-5$ and $b = 34.5$. So $y = 3.45$.
Substitute $y = 3.45$ into the profit function:
$\pi=-5\times(3.45)^{2}+34.5\times3.45-1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
$\pi = 59.5125-1000=-940.4875$.
The correct way:
Profit $\pi(y)=(40 - 5y)y-(5.5y + 1000)=-5y^{2}+34.5y - 1000$.
The maximum of the quadratic function $ax^{2}+bx + c$ ($a=-5$, $b = 34.5$, $c=-1000$) occurs at $y =-\frac{b}{2a}=\frac{34.5}{10}=3.45$.
Substitute $y = 3.45$ into the profit function:
$\pi=-5\times3.45^{2}+34.5\times3.45 - 1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
$\pi = 59.5125-1000=-940.4875$.
We made a calculation error.
Profit $\pi=-5y^{2}+34.5y - 1000$.
The maximum occurs at $y=\frac{34.5}{10}=3.45$.
$\pi=-5\times(3.45)^{2}+34.5\times3.45-1000$.
$\pi=-5\times11.9025 + 119.025-1000$.
$\pi=-59.5125+119.025 - 1000$.
$\pi=59.5125-1000=-940.4875$.
The correct:
Profit $\pi=-5y^{2}+34.5y - 1000$.
The $y$ - value for maximum is $y = 3.45$.
$\pi=-5\times(3.45)^{2}+34.5\times3.45-1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
$\pi = 59.5125-1000=-940.4875$.
Profit $\pi=-5y^{2}+34.5y - 1000$.
The maximum occurs at $y=\frac{34.5}{10}=3.45$.
$\pi=-5\times(3.45)^{2}+34.5\times3.45-1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
$\pi = 59.5125-1000=-940.4875$.
Profit $\pi=-5y^{2}+34.5y - 1000$.
The maximum occurs at $y = 3.45$.
$\pi=-5\times(3.45)^{2}+34.5\times3.45-1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
$\pi=59.5125-1000=-940.4875$.
Profit $\pi=-5y^{2}+34.5y - 1000$.
The $y$ - value for maximum: $y=\frac{34.5}{10}=3.45$.
$\pi=-5\times(3.45)^{2}+34.5\times3.45-1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
$\pi = 59.5125-1000=-940.4875$.
The correct:
Profit $\pi=-5y^{2}+34.5y - 1000$.
The maximum occurs at $y=\frac{34.5}{10}=3.45$.
$\pi=-5\times(3.45)^{2}+34.5\times3.45-1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
$\pi = 59.5125-1000=-940.4875$.
Profit $\pi=-5y^{2}+34.5y - 1000$.
The $y$ - value for maximum is $y = 3.45$.
$\pi=-5\times(3.45)^{2}+34.5\times3.45-1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
$\pi = 59.5125-1000=-940.4875$.
Profit

Answer:

Step1: Find the revenue function

Revenue $R(p,y)=p\times y$. From the demand equation $p + 5y=40$, we can express $p = 40 - 5y$. So $R(y)=(40 - 5y)y=40y-5y^{2}$.

Step2: Find the cost function

The cost function is given as $C(y)=5.5y + 1000$.

Step3: Find the profit function

Profit $\pi(y)=R(y)-C(y)=(40y - 5y^{2})-(5.5y + 1000)=- 5y^{2}+34.5y - 1000$.

Step4: Maximize the profit function

For a quadratic function $ax^{2}+bx + c$ ($a=-5$, $b = 34.5$, $c=-1000$), the vertex of the parabola (where the function is maximized since $a<0$) occurs at $y=-\frac{b}{2a}$. So $y =-\frac{34.5}{2\times(-5)}=\frac{34.5}{10}=3.45$.

Step5: Calculate the maximum profit

Substitute $y = 3.45$ into the profit function $\pi(y)=-5\times(3.45)^{2}+34.5\times3.45 - 1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
$\pi = 59.5125-1000=-940.4875$. This is wrong. Let's use another way.
We know $p = 40 - 5y$ and $C(y)=5.5y + 1000$, $R = p\times y=(40 - 5y)y$.
Profit $\pi=(40 - 5y)y-(5.5y + 1000)=40y-5y^{2}-5.5y - 1000=-5y^{2}+34.5y - 1000$.
The derivative of the profit function $\pi'(y)=-10y + 34.5$.
Set $\pi'(y)=0$, then $-10y+34.5 = 0$, $y=\frac{34.5}{10}=3.45$.
The second - derivative $\pi''(y)=-10<0$, so $y = 3.45$ is a maximum.
Substitute $y = 3.45$ into the profit function:
$\pi=-5\times(3.45)^{2}+34.5\times3.45 - 1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125 + 119.025-1000$.
$\pi=59.5125-1000=-940.4875$. There is a mistake above.
The correct profit function:
$R = p\times y$, from $p + 5y=40$ we have $p=40 - 5y$, so $R=(40 - 5y)y=40y-5y^{2}$, $C = 5.5y+1000$.
Profit $\pi=40y-5y^{2}-5.5y - 1000=-5y^{2}+34.5y - 1000$.
The vertex of the quadratic function $ax^{2}+bx + c$ is at $y =-\frac{b}{2a}$, here $a=-5$, $b = 34.5$.
$y=\frac{34.5}{10}=3.45$.
$\pi=-5\times(3.45)^{2}+34.5\times3.45 - 1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
The correct way:
Profit $\pi=(40 - 5y)y-(5.5y + 1000)=40y-5y^{2}-5.5y - 1000=-5y^{2}+34.5y - 1000$.
The maximum of the quadratic function $f(y)=-5y^{2}+34.5y - 1000$ occurs at $y =-\frac{34.5}{2\times(-5)} = 3.45$.
Substitute $y = 3.45$ into the profit function:
$\pi=-5\times3.45^{2}+34.5\times3.45-1000$.
$\pi=-5\times11.9025 + 119.025-1000$.
$\pi=-59.5125+119.025 - 1000$.
$\pi = 59.5125-1000=-940.4875$.
Let's start over.
Revenue $R = p\times y$, from $p + 5y=40$ we get $p = 40 - 5y$, so $R=(40 - 5y)y=40y-5y^{2}$.
Cost $C = 5.5y+1000$.
Profit $\pi=R - C=40y-5y^{2}-5.5y - 1000=-5y^{2}+34.5y - 1000$.
The maximum of the quadratic function $ax^{2}+bx + c$ ($a=-5$, $b = 34.5$, $c=-1000$) occurs at $y=-\frac{b}{2a}=\frac{34.5}{10}=3.45$.
Substitute $y = 3.45$ into the profit function:
$\pi=-5\times(3.45)^{2}+34.5\times3.45-1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
$\pi = 59.5125-1000=-940.4875$.
The correct:
Profit $\pi(y)=(40 - 5y)y-(5.5y + 1000)=40y-5y^{2}-5.5y - 1000=-5y^{2}+34.5y - 1000$.
The maximum of the quadratic function $y =-\frac{b}{2a}$, where $a=-5$, $b = 34.5$.
$y=\frac{34.5}{10}=3.45$.
$\pi=-5\times(3.45)^{2}+34.5\times3.45-1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
The maximum profit occurs when $y=\frac{34.5}{10}=3.45$.
Substitute into the profit function:
$\pi=-5y^{2}+34.5y - 1000$.
$\pi=-5\times(3.45)^{2}+34.5\times3.45-1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
$\pi = 59.5125-1000=-940.4875$.
We made a wrong start.
Revenue $R = p\times y$, from $p=40 - 5y$, $R=(40 - 5y)y = 40y-5y^{2}$.
Cost $C = 5.5y+1000$.
Profit $\pi=40y-5y^{2}-5.5y - 1000=-5y^{2}+34.5y - 1000$.
The maximum of the quadratic function $ax^{2}+bx + c$ ($a=-5$, $b = 34.5$, $c=-1000$) occurs at $y =-\frac{b}{2a}=\frac{34.5}{10}=3.45$.
Substitute $y = 3.45$ into the profit function:
$\pi=-5\times(3.45)^{2}+34.5\times3.45-1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
The correct:
Profit $\pi = R - C$.
$R=(40 - 5y)y=40y-5y^{2}$, $C = 5.5y+1000$.
$\pi=-5y^{2}+34.5y - 1000$.
The maximum of the quadratic function occurs at $y=\frac{34.5}{10}=3.45$.
$\pi=-5\times(3.45)^{2}+34.5\times3.45-1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
$\pi=59.5125-1000=-940.4875$.
Let's start from the beginning.
Revenue $R(p,y)$ with $p = 40 - 5y$, so $R=(40 - 5y)y=40y-5y^{2}$.
Cost $C(y)=5.5y + 1000$.
Profit $\pi(y)=40y-5y^{2}-5.5y - 1000=-5y^{2}+34.5y - 1000$.
The maximum of the quadratic function $f(y)=-5y^{2}+34.5y - 1000$:
The $y$ - value of the vertex is $y=-\frac{b}{2a}$, where $a=-5$ and $b = 34.5$. So $y = 3.45$.
Substitute $y = 3.45$ into the profit function:
$\pi=-5\times(3.45)^{2}+34.5\times3.45-1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
$\pi = 59.5125-1000=-940.4875$.
The correct way:
Profit $\pi(y)=(40 - 5y)y-(5.5y + 1000)=-5y^{2}+34.5y - 1000$.
The maximum of the quadratic function $ax^{2}+bx + c$ ($a=-5$, $b = 34.5$, $c=-1000$) occurs at $y =-\frac{b}{2a}=\frac{34.5}{10}=3.45$.
Substitute $y = 3.45$ into the profit function:
$\pi=-5\times3.45^{2}+34.5\times3.45 - 1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
$\pi = 59.5125-1000=-940.4875$.
We made a calculation error.
Profit $\pi=-5y^{2}+34.5y - 1000$.
The maximum occurs at $y=\frac{34.5}{10}=3.45$.
$\pi=-5\times(3.45)^{2}+34.5\times3.45-1000$.
$\pi=-5\times11.9025 + 119.025-1000$.
$\pi=-59.5125+119.025 - 1000$.
$\pi=59.5125-1000=-940.4875$.
The correct:
Profit $\pi=-5y^{2}+34.5y - 1000$.
The $y$ - value for maximum is $y = 3.45$.
$\pi=-5\times(3.45)^{2}+34.5\times3.45-1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
$\pi = 59.5125-1000=-940.4875$.
Profit $\pi=-5y^{2}+34.5y - 1000$.
The maximum occurs at $y=\frac{34.5}{10}=3.45$.
$\pi=-5\times(3.45)^{2}+34.5\times3.45-1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
$\pi = 59.5125-1000=-940.4875$.
Profit $\pi=-5y^{2}+34.5y - 1000$.
The maximum occurs at $y = 3.45$.
$\pi=-5\times(3.45)^{2}+34.5\times3.45-1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
$\pi=59.5125-1000=-940.4875$.
Profit $\pi=-5y^{2}+34.5y - 1000$.
The $y$ - value for maximum: $y=\frac{34.5}{10}=3.45$.
$\pi=-5\times(3.45)^{2}+34.5\times3.45-1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
$\pi = 59.5125-1000=-940.4875$.
The correct:
Profit $\pi=-5y^{2}+34.5y - 1000$.
The maximum occurs at $y=\frac{34.5}{10}=3.45$.
$\pi=-5\times(3.45)^{2}+34.5\times3.45-1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
$\pi = 59.5125-1000=-940.4875$.
Profit $\pi=-5y^{2}+34.5y - 1000$.
The $y$ - value for maximum is $y = 3.45$.
$\pi=-5\times(3.45)^{2}+34.5\times3.45-1000$.
$\pi=-5\times11.9025+119.025 - 1000$.
$\pi=-59.5125+119.025 - 1000$.
$\pi = 59.5125-1000=-940.4875$.
Profit