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3 years ago

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You want to buy a car, and a local bank will lend you $30,000. The loan would be fully amortized over 6 years (72 months), and t

he nominal interest rate would be 6%, with interest paid monthly. What is the monthly loan payment? Do not round intermediate calculations. Round your answer to the nearest cent.
$

What is the loan's EFF%? Do not round intermediate calculations. Round your answer to two decimal places.

%

1 answer:

Georgia [21]3 years ago

6 0

Answer:

<h2>- 56,667cents</h2><h2>- 6.17%</h2>

Step-by-step explanation:

Before we can determine the monthly loan paymnet, we must first calculate the total amount paid at the end of 6years.

Amount = Principal + Interest

Given Principal = $30,000

Interest = Principal * rate * time/100

Interest = $30,000*6*6/100

Interest = $10,800

Amount = $30,000+ $10,800

Amount = $40,800

If amount paid after 6years is $40,800, my monthly loan payment = $40,800/72 ≈ $566.67 to nearest dollar.

since $1 - 100cents

$566.67 = 100 * 566.67

$567 = 56,667cents

Monthly loan payment to nearest cent will be 56,667cents

EFF = (1 + r /n)^n - 1

r is the rate and n is the number of period per year which is 12months

%EFF = EAR = (1 + 0.06 /12)^ 12 - 1

%EFF = 1.005^12

%EFF = 1.061678 - 1

%EFF = 0.061678

%EFF = 6.17% to 2dp

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Step-by-step explanation:

<u>Optimizing With Lagrange Multipliers</u>

When a multivariable function f is to be maximized or minimized, the Lagrange multipliers method is a pretty common and easy tool to apply when the restrictions are in the form of equalities.

Consider three generators that can output xi megawatts, with i ranging from 1 to 3. The set of unknown variables is x1, x2, x3.

The cost of each generator is given by the formula

$\displaystyle C_i=3x_i+\frac{i}{40}x_i^2$

It means the cost for each generator is expanded as

$\displaystyle C_1=3x_1+\frac{1}{40}x_1^2$

$\displaystyle C_2=3x_2+\frac{2}{40}x_2^2$

$\displaystyle C_3=3x_3+\frac{3}{40}x_3^2$

The total cost of production is

$\displaystyle C(x_1,x_2,x_3)=3x_1+\frac{1}{40}x_1^2+3x_2+\frac{2}{40}x_2^2+3x_3+\frac{3}{40}x_3^2$

Simplifying and rearranging, we have the objective function to minimize:

$\displaystyle C(x_1,x_2,x_3)=3(x_1+x_2+x_3)+\frac{1}{40}(x_1^2+2x_2^2+3x_3^2)$

The restriction can be modeled as a function g(x)=0:

$g: x_1+x_2+x_3=1000$

Or

$g(x_1,x_2,x_3)= x_1+x_2+x_3-1000$

We now construct the auxiliary function

$f(x_1,x_2,x_3)=C(x_1,x_2,x_3)-\lambda g(x_1,x_2,x_3)$

$\displaystyle f(x_1,x_2,x_3)=3(x_1+x_2+x_3)+\frac{1}{40}(x_1^2+2x_2^2+3x_3^2)-\lambda (x_1+x_2+x_3-1000)$

We find all the partial derivatives of f and equate them to 0

$\displaystyle f_{x1}=3+\frac{2}{40}x_1-\lambda=0$

$\displaystyle f_{x2}=3+\frac{4}{40}x_2-\lambda=0$

$\displaystyle f_{x3}=3+\frac{6}{40}x_3-\lambda=0$

$f_\lambda=x_1+x_2+x_3-1000=0$

Solving for \lambda in the three first equations, we have

$\displaystyle \lambda=3+\frac{2}{40}x_1$

$\displaystyle \lambda=3+\frac{4}{40}x_2$

$\displaystyle \lambda=3+\frac{6}{40}x_3$

Equating them, we find:

$x_1=3x_3$

$\displaystyle x_2=\frac{3}{2}x_3$

Replacing into the restriction (or the fourth derivative)

$x_1+x_2+x_3-1000=0$

$\displaystyle 3x_3+\frac{3}{2}x_3+x_3-1000=0$

$\displaystyle \frac{11}{2}x_3=1000$

$x_3=181.8\ MW$

And also

$x_1=545.5\ MW$

$x_2=272.7\ MW$

The load balance $(x_1,x_2,x_3)=(545.5,272.7,181.8)$ Mw minimizes the total cost

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3 years ago