the demand for grass seed (in thousands of pounds) at price p dollars is given by the following function…

the demand for grass seed (in thousands of pounds) at price p dollars is given by the following function. d(p)= - 3p³ - 2p² + 1521 use the differential to approximate the changes in demand for the following changes in p. a. $4 to $4.12 b. $5 to $5.13 a. the change in demand based on a change in p from $4 to $4.12 is about thousand pounds. (type an integer or a decimal.)

the demand for grass seed (in thousands of pounds) at price p dollars is given by the following function. d(p)= - 3p³ - 2p² + 1521 use the differential to approximate the changes in demand for the following changes in p. a. $4 to $4.12 b. $5 to $5.13 a. the change in demand based on a change in p from $4 to $4.12 is about thousand pounds. (type an integer or a decimal.)

Answer

Explanation:

Step1: Find the derivative of D(p)

First, find the derivative of $D(p)=- 3p^{3}-2p^{2}+1521$. Using the power - rule $\frac{d}{dp}(x^{n})=nx^{n - 1}$, we have $D^\prime(p)=-9p^{2}-4p$.

Step2: Calculate $\Delta p$ for part a

For the change from $p = 4$ to $p = 4.12$, $\Delta p=4.12 - 4=0.12$.

Step3: Evaluate $D^\prime(p)$ at $p = 4$

Substitute $p = 4$ into $D^\prime(p)$: $D^\prime(4)=-9\times4^{2}-4\times4=-9\times16 - 16=-144 - 16=-160$.

Step4: Approximate the change in demand for part a

The differential $dD\approx D^\prime(p)\Delta p$. Substitute $D^\prime(4)=-160$ and $\Delta p = 0.12$ into the formula, we get $dD\approx-160\times0.12=-19.2$.

Step5: Calculate $\Delta p$ for part b

For the change from $p = 5$ to $p = 5.13$, $\Delta p=5.13 - 5=0.13$.

Step6: Evaluate $D^\prime(p)$ at $p = 5$

Substitute $p = 5$ into $D^\prime(p)$: $D^\prime(5)=-9\times5^{2}-4\times5=-9\times25-20=-225 - 20=-245$.

Step7: Approximate the change in demand for part b

The differential $dD\approx D^\prime(p)\Delta p$. Substitute $D^\prime(5)=-245$ and $\Delta p = 0.13$ into the formula, we get $dD\approx-245\times0.13=-31.85$.

Answer:

a. - 19.2 b. - 31.85