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3 years ago

If $450 is invested for one year at 55% simple interest, how much interest is earned?

Mathematics

2 answers:

Aleonysh [2.5K]3 years ago

5 0

I think it is 550. 65

BartSMP [9]3 years ago

3 0

Answer:

$247.50

Step-by-step explanation:

You didn't specify, so I am going to assume that it is annual, or yearly, interest.

You want to calculate the interest on $450 at 55% interest per year after 1 year.

The formula we'll use for this is the simple interest formula, or:

I= P × r × t

Where...

P equals the principal amount, which is $450.00.

r equals the interest rate, which would be 55% per year, or as a decimal, 55/100=0.55.

t equals time involved, which is 1 year

To find the simple interest, we multiply 450 × 0.55 × 1 to get that the interest is $247.50

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Let x be the width and y the length of the rectangular field.

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We have to figure out where the function is increasing and decreasing. Differentiating,

$\frac{d}{dx}C=\frac{d}{dx}\left(20x+\frac{18000}{x}\right)\\\\C'=20-\frac{18000}{x^2}$

Next, we find the critical points of the derivative

$20-\frac{18000}{x^2}=0\\\\20x^2-\frac{18000}{x^2}x^2=0\cdot \:x^2\\\\20x^2-18000=0\\\\20x^2-18000+18000=0+18000\\\\20x^2=18000\\\\\frac{20x^2}{20}=\frac{18000}{20}\\\\x^2=900\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{900},\:x=-\sqrt{900}\\\\x=30,\:x=-30$

Because the length is always positive the only point we take is $x=30$ . We thus test the intervals $(0, 30)$ and $(30, \infty)$

$C'(20)=20-\frac{18000}{20^2} = -25 < 0\\\\C'(40)= 20-\frac{18000}{20^2} = 8.75 >0$

we see that total cost function is decreasing on $(0, 30)$ and increasing on $(30, \infty)$ . Therefore, the minimum is attained at $x=30$ , so the minimal cost is

$C(30)=20(30)+\frac{18000}{30}\\C(30)=1200$