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

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

Step-by-step explanation:

The revenue function is

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.

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

Divide both sides by 3.

Taking square root both the sides.

Find double derivative of the function.

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.

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°