Answer :
Let's address the given problem and solve it step by step:
### Question 4:
An enterprise produces computer terminals and after analyzing its production, it is determined that the profit in dollars, [tex]\( P(x) \)[/tex], when selling [tex]\( x \)[/tex] terminals per month, is given by:
[tex]\[ P(x) = 0.1x^2 - 160x - 200 \][/tex]
#### Part (a)
Question: How many terminals should be sold per month to achieve maximum profit?
To determine the number of terminals that should be sold to achieve maximum profit, we need to find the vertex of the quadratic function [tex]\( P(x) = 0.1x^2 - 160x - 200 \)[/tex].
The form of the quadratic function is [tex]\( ax^2 + bx + c \)[/tex], and the vertex can be found using the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
In this function:
- [tex]\( a = 0.1 \)[/tex]
- [tex]\( b = -160 \)[/tex]
Now, using the formula:
[tex]\[ x = -\frac{-160}{2 \times 0.1} \][/tex]
[tex]\[ x = \frac{160}{0.2} \][/tex]
[tex]\[ x = 800 \][/tex]
So, the company should sell 800 terminals per month to achieve the maximum profit.
#### Part (b)
Question: What is the maximum profit?
To find the maximum profit, we substitute [tex]\( x = 800 \)[/tex] into the profit function [tex]\( P(x) \)[/tex]:
[tex]\[ P(800) = 0.1 \times (800)^2 - 160 \times 800 - 200 \][/tex]
First, calculate [tex]\( (800)^2 \)[/tex]:
[tex]\[ (800)^2 = 640000 \][/tex]
Now, substitute this value in:
[tex]\[ P(800) = 0.1 \times 640000 - 160 \times 800 - 200 \][/tex]
[tex]\[ P(800) = 64000 - 128000 - 200 \][/tex]
[tex]\[ P(800) = 64000 - 128000 - 200 \][/tex]
[tex]\[ P = 64000 - 128000 \][/tex]
[tex]\[ P = -64000 - 200 \][/tex]
[tex]\[ P = -64200 \][/tex]
Therefore, the maximum profit is [tex]$ -64,200. Thus, the company should sell 800 terminals per month to achieve the utility maximum, which would amount to a maximum profit of $[/tex] -64,200.
### Summary:
a) The number of terminals to be sold per month for maximum profit: 800 terminals.
b) The maximum profit: $ -64,200.
### Question 4:
An enterprise produces computer terminals and after analyzing its production, it is determined that the profit in dollars, [tex]\( P(x) \)[/tex], when selling [tex]\( x \)[/tex] terminals per month, is given by:
[tex]\[ P(x) = 0.1x^2 - 160x - 200 \][/tex]
#### Part (a)
Question: How many terminals should be sold per month to achieve maximum profit?
To determine the number of terminals that should be sold to achieve maximum profit, we need to find the vertex of the quadratic function [tex]\( P(x) = 0.1x^2 - 160x - 200 \)[/tex].
The form of the quadratic function is [tex]\( ax^2 + bx + c \)[/tex], and the vertex can be found using the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
In this function:
- [tex]\( a = 0.1 \)[/tex]
- [tex]\( b = -160 \)[/tex]
Now, using the formula:
[tex]\[ x = -\frac{-160}{2 \times 0.1} \][/tex]
[tex]\[ x = \frac{160}{0.2} \][/tex]
[tex]\[ x = 800 \][/tex]
So, the company should sell 800 terminals per month to achieve the maximum profit.
#### Part (b)
Question: What is the maximum profit?
To find the maximum profit, we substitute [tex]\( x = 800 \)[/tex] into the profit function [tex]\( P(x) \)[/tex]:
[tex]\[ P(800) = 0.1 \times (800)^2 - 160 \times 800 - 200 \][/tex]
First, calculate [tex]\( (800)^2 \)[/tex]:
[tex]\[ (800)^2 = 640000 \][/tex]
Now, substitute this value in:
[tex]\[ P(800) = 0.1 \times 640000 - 160 \times 800 - 200 \][/tex]
[tex]\[ P(800) = 64000 - 128000 - 200 \][/tex]
[tex]\[ P(800) = 64000 - 128000 - 200 \][/tex]
[tex]\[ P = 64000 - 128000 \][/tex]
[tex]\[ P = -64000 - 200 \][/tex]
[tex]\[ P = -64200 \][/tex]
Therefore, the maximum profit is [tex]$ -64,200. Thus, the company should sell 800 terminals per month to achieve the utility maximum, which would amount to a maximum profit of $[/tex] -64,200.
### Summary:
a) The number of terminals to be sold per month for maximum profit: 800 terminals.
b) The maximum profit: $ -64,200.
