Question

Suppose a revenue function is given by: R ( q ) = − q 3 + 140 q where q is thousands of units and R ( q ) is thousands of dollars. For what value of q is revenue maximized. Round your answer to the nearest tenth (one decimal place). q = Incorrect thousand units

Answer 1

Answer:

At q=6.8 the revenue is maximum. So, q=6.8 thousand units.

Step-by-step explanation:

The revenue function is

$R(q)=-q^3+140q$

where q is thousands of units and R ( q ) is thousands of dollars.

We need to find for what value of q is revenue maximized.

Differentiate the function with respect to q.

$R'(q)=-3q^2+140$

Equate R'(q)=0, to find the critical values.

$0=-3q^2+140$

$3q^2=140$

Divide both sides by 3.

$q^2=\frac{140}{3}$

Taking square root both the sides.

$q=\pm \sqrt{\frac{140}{3}}$

$q=\pm 6.8313$

$q\approx \pm 6.8$

Find double derivative of the function.

$R''(q)=-6q$

For q=-6.8, R''(q)>0 and q=6.8, R''(q)<0. So at q=6.8 revenue is maximum.

At q=6.8 the revenue is maximum. So, q=6.8 thousand units.