Let's go step by step for the given demand function: **Demand function:** \[ q = D(x) = 2x + \frac{400}{20x + 7} \] --- ### (a) **Find the elasticity** **Elasticity of demand** \( E(x) \) is given by: \[ E(x) = -\frac{x}{q} \cdot \frac{dq}{dx} \] Let's compute \( \frac{dq}{dx} \): \[ q = 2x + \frac{400}{20x + 7} \] Let’s differentiate \( q \) with respect to \( x \): - The derivative of \( 2x \) is \( 2 \). - The derivative of \( \frac{400}{20x+7} \) (use the quotient or chain rule): Let \( u = 400 \), \( v = 20x+7 \): \[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \] Since \( u' = 0 \) and \( v' = 20 \): \[ \frac{d}{dx}\left(\frac{400}{20x+7}\right) = \frac{0 \cdot (20x+7) - 400 \cdot 20}{(20x+7)^2} = -\frac{8000}{(20x+7)^2} \] So, \[ \frac{dq}{dx} = 2 - \frac{8000}{(20x+7)^2} \] Now plug into the elasticity formula: \[ E(x) = -\frac{x}{q} \cdot \left(2 - \frac{8000}{(20x+7)^2}\right) \] --- ### (b) **Find the elasticity when \( x = 3 \)** First, compute \( q \) at \( x = 3 \): \[ q = 2(3) + \frac{400}{20(3)+7} = 6 + \frac{400}{60+7} = 6 + \frac{400}{67} \approx 6 + 5.97 \approx 11.97 \] Now, compute \( (20x+7)^2 \) at \( x = 3 \): \[ 20x+7 = 67 \\ (20x+7)^2 = 67^2 = 4489 \] Now, \( \frac{dq}{dx} \) at \( x = 3 \): \[ \frac{dq}{dx} = 2 - \frac{8000}{4489} \approx 2 - 1.782 = 0.218 \] Plug these into the elasticity formula: \[ E(3) = -\frac{3}{11.97} \cdot 0.218 \approx -0.250 \cdot 0.218 \approx -0.0545 \] --- ### (c) **At $3 per plant, will a small increase in price cause total revenue to increase or decrease?** If \( |E(x)| < 1 \), demand is **inelastic** and an increase in price increases total revenue. Here, \[ |E(3)| \approx 0.0545 < 1 \] So, **demand is inelastic** at \( x = 3 \). **Therefore, a small increase in price will cause the total revenue to increase.** --- ## **Final Answers** **(a) The elasticity is:** \[ E(x) = -\frac{x}{2x + \frac{400}{20x+7}}\left(2 - \frac{8000}{(20x+7)^2}\right) \] **(b) The elasticity at \( x = 3 \) is:** \[ E(3) \approx -0.0545 \] **(c) At $3 per plant, a small increase in price will cause the total revenue to increase.**