# Here is a very long maths problem. Please help...The F.Bender Insurance company is planning to offer a senior citizen's accidents policy. From previous records the agent knows that the number of...

# Here is a very long maths problem. Please help...

The F.Bender Insurance company is planning to offer a senior citizen's accidents policy. From previous records the agent knows that the number of accidents per 100 million kilometers driven in terms of age can be modelled by the quadratic function y=0.4x^2 - 36x + 1000 where x is the driver's age and y is the number of accidents. If the company decides to insure drivers up to the age where the accident rate becomes 830 accident per 100 kilometers, what could be the age of their oldest driver?a)64 b)72 c)79 d)85 e)90.

I got my answer to be close to D but I am not sure if it is correct. Can anyone help please?

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### 2 Answers

Here we are given the number of accidents as y=0.4x^2 - 36x + 1000, where x is the driver's age. As y can take the maximum value of 830, let's equate 0.4x^2 - 36x + 1000 to 830 and solve the resulting equation.

So we have 0.4x^2 - 36x + 1000 = 830

=> 0.4x^2 - 36x + 170 = 0

Divide by 0.4

=> x^2 - 90x + 425 = 0

=> x^2 - 85x - 5x + 425 = 0

=> x(x - 85) -5(x- 85) = 0

=> (x-5) (x-85) = 0

For x-5 = 0, we have x=5

and for x-85 = 0, we have x= 85.

As we need to find the age of the oldest person who can be insured, we take x = 85.

**So, the right option is d.**

y=0.4x^2 - 36x + 1000

We need to determine the maximum age for the driver,

From the formula, we can obtain the minimum value by differentiate the function y and find the critical value.

Then the minimum value will represent the minimum accidentsÂ for the driver.

Let us differentiate:

y=0.4x^2 - 36x + 1000

==> y' = 2*0.4x^2 - 36

==> y'= 0.8x - 36

==> 0.8 x = 36

==> x= 36/0.8 = 45