question 10 7 pts solve the problem. $1400 is deposited in an account with an interest rate of r % per year…

question 10 7 pts solve the problem. $1400 is deposited in an account with an interest rate of r % per year, compounded monthly. at the end of 8 years, the balance in the account is given by a = 1400(1 + r/1200)^96. find the rate of change of a with respect to r when r = 4 round answer to the nearest hundredth, if necessary. edit view insert format tools table 12pt paragraph b i u a t2

question 10 7 pts solve the problem. $1400 is deposited in an account with an interest rate of r % per year, compounded monthly. at the end of 8 years, the balance in the account is given by a = 1400(1 + r/1200)^96. find the rate of change of a with respect to r when r = 4 round answer to the nearest hundredth, if necessary. edit view insert format tools table 12pt paragraph b i u a t2

Answer

Explanation:

Step1: Recall the power - rule for differentiation

If $y = a(u)^n$, then $y^\prime=an(u)^{n - 1}u^\prime$. Let $u = 1+\frac{r}{1200}$ and $n = 96$, $a = 1400$. The derivative of $u$ with respect to $r$ is $u^\prime=\frac{1}{1200}$. The derivative of $A$ with respect to $r$ using the chain - rule is $A^\prime=1400\times96\times(1 + \frac{r}{1200})^{95}\times\frac{1}{1200}$.

Step2: Simplify the derivative formula

$A^\prime=\frac{1400\times96}{1200}(1+\frac{r}{1200})^{95}=112(1+\frac{r}{1200})^{95}$.

Step3: Substitute $r = 4$ into the derivative formula

When $r = 4$, we have $1+\frac{r}{1200}=1+\frac{4}{1200}=1+\frac{1}{300}=\frac{301}{300}$. Then $A^\prime=112\times(\frac{301}{300})^{95}$. Using a calculator, $(\frac{301}{300})^{95}\approx1.3679$ and $A^\prime=112\times1.3679 = 153.2048\approx153.20$.

Answer:

$153.20$