Question

the daily profit in dollars of a specialty cake shop is described by the function P(x)=-3x^2+168x-1920. Where x is the number of cakes prepared in one day. The maximum profit for the company occurs at the vertex of the parabola. How many cakes should be prepared per day in order to maximize profit?

Answer 1

Answer:

28

Step-by-step explanation:

We have the following function:

$f(x)=-3x^{2} +168x-1920$

where a=-3, b=168 and c=-1920

In order to calculate the maxium profit for the company, and how many cakes should be prepared in order to reach it, we have to calculate where the parabola's vertex is ubicated. To do so, we use the following formula:

$Xv= \frac{-b}{2a}$

$Xv= \frac{-168}{2.(-3)}$

$Xv= \frac{-168}{2.(-3)}$

$Xv= 28$

So 28 cakes should be prepared per day in order to maximize profit.

Answer 2

Answer:

Number of cakes should be prepared per day in order to maximize profit is 28.

Step-by-step explanation:

Given : The daily profit in dollars of a specialty cake shop is described by the function $P(x)=-3x^2+168x-1920$ where x is the number of cakes prepared in one day.

To find : How many cakes should be prepared per day in order to maximize profit?

Solution :

The maximum profit for the company occurs at the vertex of the parabola.

So, the required vertex of a quadratic equation $ax^2+bx+c$ is given by

$x=-\frac{b}{2a}$

On comparing the profit function,

a=-3, b=168 and c=-1920

Substitute,

$x=-\frac{168}{2(-3)}$

$x=\frac{168}{6}$

$x=28$

So, number of cakes should be prepared per day in order to maximize profit is 28.