College Algebra

Linear Algebra

To find the elasticity of demand in terms of \( p \), we first need to express the demand equation in the form \( q = f(p) \). The given demand equation is: \[ q^2(3 + p^3) = 44 \] Let's solve for \( q \) in terms of \( p \): \[ q^2 = \frac{44}{3 + p^3} \] Taking the square root of both sides, we get: \[ q = \sqrt{\frac{44}{3 + p^3}} \] Now, the price elasticity of demand (E) is defined as: \[ E = \frac{d q}{d p} \cdot \frac{p}{q} \] Let's find \( \frac{d q}{d p} \): \[ \frac{d q}{d p} = \frac{d}{d p} \left( \frac{44}{3 + p^3} \right)^{1/2} \] Using the chain rule, we have: \[ \frac{d q}{d p} = \frac{1}{2} \left( \frac{44}{3 + p^3} \right)^{-1/2} \cdot \left( -\frac{44 \cdot 3p^2}{(3 + p^3)^2} \right) \] Simplifying, we get: \[ \frac{d q}{d p} = -\frac{22p^2}{(3 + p^3)^{3/2}} \] Now, we can plug \( \frac{d q}{d p} \) and \( q \) into the elasticity formula: \[ E = \left( -\frac{22p^2}{(3 + p^3)^{3/2}} \right) \cdot \frac{p}{\sqrt{\frac{44}{3 + p^3}}} \] Simplifying further: \[ E = -\frac{22p^3}{(3 + p^3)^2} \cdot \frac{(3 + p^3)^{1/2}}{\sqrt{44}} \] \[ E = -\frac{22p^3}{(3 + p^3)^{3/2}} \cdot \frac{(3 + p^3)^{1/2}}{\sqrt{44}} \] \[ E = -\frac{22p^3}{44} \cdot \frac{1}{3 + p^3} \] \[ E = -\frac{p^3}{2} \cdot \frac{1}{3 + p^3} \] \[ E = -\frac{p^3}{2(3 + p^3)} \] To show that \( E \) never exceeds \( \frac{3}{2} \), we need to show that: \[ -\frac{p^3}{2(3 + p^3)} \leq \frac{3}{2} \] Since \( E \) is negative (indicating that the demand is elastic), we only need to consider the magnitude of \( E \) without the negative sign: \[ \frac{p^3}{2(3 + p^3)} \leq \frac{3}{2} \] Multiplying both sides by \( 2(3 + p^3) \), we get: \[ p^3 \leq 3(3 + p^3) \] \[ p^3 \leq 9 + 3p^3 \] Subtracting \( p^3 \) from both sides: \[ 0 \leq 9 + 2p^3 \] Since \( p^3 \) is always non-negative, the inequality holds true for all \( p \), and thus \( E \) never exceeds \( \frac{3}{2} \). To find for what values of \( p \) the revenue from this market is a decreasing function of \( p \), we need to consider the revenue function \( R(p) \), which is: \[ R(p) = p \cdot q \] \[ R(p) = p \cdot \sqrt{\frac{44}{3 + p^3}} \] The revenue is a decreasing function of \( p \) when \( \frac{d R}{d p} < 0 \). Let's find \( \frac{d R}{d p} \): \[ \frac{d R}{d p} = \sqrt{\frac{44}{3 + p^3}} + p \cdot \frac{d}{d p} \left( \sqrt{\frac{44}{3 + p^3}} \right) \] We already found \( \frac{d q}{d p} \) earlier, so we can use it here: \[ \frac{d R}{d p} = \sqrt{\frac{44}{3 + p^3}} - \frac{22p^3}{(3 + p^3)^{3/2}} \] For \( \frac{d R}{d p} \) to be negative, the following must hold: \[ \sqrt{\frac{44}{3 + p^3}} < \frac{22p^3}{(3 + p^3)^{3/2}} \] Squaring both sides to eliminate the square root, we get: \[ \frac{44}{3 + p^3} < \frac{484p^6}{(3 + p^3)^3} \] Multiplying both sides by \( (3 + p^3)^3 \), we obtain: \[ 44(3 + p^3)^2 < 484p^6 \] This inequality will give us the range of \( p \) values for which the revenue is a decreasing function of \( p \). Solving this inequality will require finding the roots of the resulting polynomial, which may not be possible to express in simple terms. However, numerical methods or graphing can be used to approximate the values of \( p \) that satisfy the inequality.

Math

Given that a market’s demand equation is \( q^2(3 + p^3) = 44 \), find the elasticity of demand in terms of \( p \) and show that it never exceeds \( \frac{3}{2} \). For what values of \( p \) is the revenue from this market a decreasing function of \( p \)?

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