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andrew-mc [135]
3 years ago
12

The price p (in dollars) that a camera manufacturer is able to charge for a camera is given by p = 100 - 10x² where x is the num

ber (in millions) of cameras produced. It costs the company $30 to make a camera. Currently, the company produces 2 million cameras and makes a profit of $60,000,000. Write and solve an equation to find other values of x that yield a profit of $60,000,000
Mathematics
2 answers:
Serga [27]3 years ago
8 0

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)

60 = 100x - 10x^3 -30(x)\\10x^3+30x-100x+60=0\\10x^3-70x+60=0\\

Dividing all through by 10

x^3-7x+6=0

When x=1, the LHS=0

So x-1 is a factor of the polynomial.

x^3-7x+6=(x-1)(x^2+x-6)=0

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

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)

Alla [95]3 years ago
4 0

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.

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