Respuesta :
Answer:
10x^3-70x+60=0
x=1, x=2.
Step-by-step explanation:
x is the number (in millions) of cameras produced
The price is p=100 - 10x²
Profit= 60 (in millions)
It costs 30 to make a camera.
So we have:
Profit=x(price-cost)
60=x( [100-10x²] - 30)
60=100x-10x^3 -30x
60=70x-10x^3
10x^3-70x+60=0
⇒x^3 -7x+6=0 is the equation we were looking for.
Solving:
The independent term of the equation is 6.
the divisors of 6 are {1,-1, 2,-2, 3,-3, 6,-6}
Substituting each divisor in the equation we have:
x=1
(1)^3 -7(1)+6=0
1-7+6=0
0=0
⇒x=1 is a solution.
x=-1
(-1)^3 -7(-1)+6=0
-1+7+6=0
12=0 this is not true
⇒x=1 is not a solution.
x=2
(2)^3 -7(2)+6=0
8-14+6=0
0=0
⇒x=2 is a solution.
x=-2
(-2)^3 -7(-2)+6=0
-8+14+6=0
12=0 this is not true
⇒x=-2 is not a solution.
x=3
(3)^3 -7(3)+6=0
27-21+6=0
12=0 this is not true
⇒x=2 is not a solution.
x=-3
(-3)^3 -7(-3)+6=0
-27+21+6=0
0=0
⇒x=-3 is a solution.
x=6
(6)^3 -7(6)+6=0
216-42+6=0 this is not true
⇒x=6 is not a solution.
x=-6
(-6)^3 -7(-6)+6=0
-216+42+6=0 this is not true
⇒x=-6 is not a solution.
The solutions of the equation are x=-3,x=1,x=2.
However, it is not possible to produce -3 millions of cameras.
So the only possible solutions are x=1 and x=2.
Answer:
x=1 (in million)
Step-by-step explanation:
Selling Price per camera , p=100 - 10x²
Cost Price Per Camera = $30
If x=2, the company makes a profit of $60,000,000
Profit= Selling Price-Cost Price
60 = x(100 - 10x²) -30(x)
[tex]60 = 100x - 10x^3 -30(x)\\10x^3+30x-100x+60=0\\10x^3-70x+60=0\\[/tex]
Dividing all through by 10
[tex]x^3-7x+6=0[/tex]
When x=1, the LHS=0
So x-1 is a factor of the polynomial.
[tex]x^3-7x+6=(x-1)(x^2+x-6)=0[/tex]
[tex]x-1=0, x^2+x-6=0\\x^2+x-6=0\\x^2+3x-2x-6=0\\x(x+3)-2(x+3)=0\\(x-2)(x+3)=0\\x=2, x=-3[/tex]
Therefore, x=1, 2, or -3
Since x cannot have a negative, the other value of x for which the firm makes a profit of $60,000,000 is 1(in million)
