after doing some market research, crunchy cookie inc. has found that their profit on cookies after spending…

after doing some market research, crunchy cookie inc. has found that their profit on cookies after spending $x$ amount on advertising can be modeled by $p(x)=5000 + 1000x-5x^{2}$. both $x$ and $p(x)$ are measured in thousands of dollars.\na. what is the most profit that crunchy cookie inc. can make?\nb. how much would they have to spend on advertising to achieve this result?\nc. is there ever a time when crunchy cookie inc. will make no profit? explain why or why not.

after doing some market research, crunchy cookie inc. has found that their profit on cookies after spending $x$ amount on advertising can be modeled by $p(x)=5000 + 1000x-5x^{2}$. both $x$ and $p(x)$ are measured in thousands of dollars.\na. what is the most profit that crunchy cookie inc. can make?\nb. how much would they have to spend on advertising to achieve this result?\nc. is there ever a time when crunchy cookie inc. will make no profit? explain why or why not.

Answer

Explanation:

Step1: Identify the function type

The profit function $P(x)=5000 + 1000x-5x^{2}$ is a quadratic function in the form $y = ax^{2}+bx + c$ where $a=-5$, $b = 1000$ and $c = 5000$.

Step2: Find the x - coordinate of the vertex

The x - coordinate of the vertex of a quadratic function $y=ax^{2}+bx + c$ is given by $x=-\frac{b}{2a}$. Substituting $a=-5$ and $b = 1000$ into the formula, we get $x=-\frac{1000}{2\times(-5)}=\frac{-1000}{-10}=100$.

Step3: Find the maximum profit

Substitute $x = 100$ into the profit function $P(x)$. $P(100)=5000+1000\times100 - 5\times(100)^{2}$ $=5000 + 100000-5\times10000$ $=5000 + 100000 - 50000$ $=55000$.

Step4: Check for zero - profit

Set $P(x)=0$, so $5000 + 1000x-5x^{2}=0$. Divide through by 5 to simplify: $1000 + 200x-x^{2}=0$, or $x^{2}-200x - 1000=0$. The discriminant of a quadratic equation $ax^{2}+bx + c=0$ is $\Delta=b^{2}-4ac$. Here, $a = 1$, $b=-200$ and $c=-1000$. $\Delta=(-200)^{2}-4\times1\times(-1000)=40000 + 4000=44000>0$. Since the discriminant is positive, there are two real - valued solutions for $x$, meaning there are times when the company will make no profit.

Answer:

a. The most profit is $55000$ thousand dollars. b. They have to spend $100$ thousand dollars on advertising to achieve this result. c. Yes, because the discriminant of the quadratic equation $x^{2}-200x - 1000 = 0$ (derived from setting $P(x)=0$) is positive ($\Delta = 44000>0$), so there are two real - valued solutions for the advertising amount $x$ when the profit is zero.