Question

I will need help with this math problem it has three parts to it

Answer 1

a)The smaller break even quantity is 10

b)The quantity that must be sold to maximize the profit= 14.25

c)The maximum profit = $36.13

STEP - BY - STEP EXPLANATION

What to find?

• The smaller break even quantity is

,

• The quantity that must be sold to maximize the profit

,

• The maximum profit.

Given:

C(x)= 13x + 370

R(x)=70x - 2x²

a) At the break even quantity, the revenue = cost, that is no profit and no loss.

R(x) = C(x)

So that we have;

70x - 2x² = 13x + 370

Re-arrange.

-2x² + 70x - 13x - 370 =0

-2x² +57x - 370 =0

2x² - 57x + 370 = 0

We can now solve the quadratic equation above.

Using the qudaratic formula

`x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

a=2 b=-57 c=370

Substitute the values into the formula and evaluate.

`x=\frac{-(-57)\pm\sqrt[]{(-57)^2-4(2)(370)}}{2(2)}`
`x=(57\pm17)/(4)`
`Either\text{ x=}(57+17)/(4)=(74)/(4)=(37)/(2)`

Or

`x=(57-17)/(4)=(40)/(4)=10`

x = 37/2 or x=10

We take the smaller value.

Hence, x=10

Therefore, the smaller break even quantity is 10.

b)To find the quantity that will maximize the profit, find the profit function.

p(x) = R(x) - C(x)

P(x) = 70x - 2x² - (13x +370)

= 70x - 2x² - 13x - 370

P(x) =-2x² + 57x - 370

Equate to zero

-2x² + 57x - 370 = 0

2x² - 57x + 370 =0

The maximum is at x = -b/2a

b= -57 and a=2

Substitute the values

x= - (-57) /2(2)

x= 57 /4

x= 14.25

c) To find the maximum profit, simply substitute x=14.25 into the profit function and simplify.

That is;

P(x) =-2x² + 57x - 370

P(14.25) =-2(14.25)² + 57(14.25) - 370

= -406.125 + 812.25 - 370

= 36.125

≈ 36.13

Hence, the maximum profit is $36.13