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iris [78.8K]
2 years ago
9

The cost to produce a product is modeled by the function f(x) = 5x2 − 70x 258, where x is the number of products produced. compl

ete the square to determine the minimum cost of producing this product. 5(x − 7)2 13; the minimum cost to produce the product is $13. 5(x − 7)2 13; the minimum cost to produce the product is $7. 5(x − 7)2 258; the minimum cost to produce the product is $7. 5(x − 7)2 258; the minimum cost to produce the product is $258.
Mathematics
1 answer:
ad-work [718]2 years ago
5 0

The minimum cost to produce the product is $7.

What are functions?

  • A function from a set X to a set Y allocates exactly one element of Y to each element of X.

To determine the minimum cost:

We have to determine the minimum cost of producing this product.

Since f(x) = 5x^{2} -70x+258.

Now, consider the equation 5x^{2} -70x+258= 0.

Dividing the above equation by 5, we get

x^{2} -70x/5+258/5=0\\x^{2} -14x+258/5=0

Now, considering the coefficient of 'x', dividing it by '2' and then adding and subtracting the square of the number which we got after dividing.

Since the coefficient of 'x' is 14, and half of 14 is '7'.

So, adding and subtracting from the above equation.

x^{2} -14x+(7)^{2} -(7)^{2}+258/5=0\\x^{2} -14x+49 -49+258/5=0\\(x-7)^{2} -49+258/5=0\\(x-7)^{2} +258-245/5=0\\(x-7)^{2} +13/5=0\\5(x-7)^{2} +13=0

Now, we have to determine the minimum cost to produce the product.

Since f(x) = 5x^{2} -70x+258.

f'(x) = 10x-70

Now, let f'(x)=0

10x-70=0\\10x=70

Therefore, x=7

Now, consider which is greater than 0.

Therefore, x= 7 is the minimum cost.

Know more about functions here:

brainly.com/question/25638609

#SPJ4

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Find the value of the function f(x) = 7x^4 - 3x^3 + 2x^2 - 8x + 10 when x = 2
lidiya [134]

Answer:

x= -25

Step-by-step explanation:

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Hey use Socratic for questions like this

3 0
2 years ago
3 3/5 + 6 3/10 please
Dafna11 [192]
In order to solve first convert both values into improper fractions:
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3 0
3 years ago
How do you find the limit?
coldgirl [10]

Answer:

2/5

Step-by-step explanation:

Hi! Whenever you find a limit, you first directly substitute x = 5 in.

\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{5^2-6(5)+5}{5^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{25-30+5}{25-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{0}{0}}

Hm, looks like we got 0/0 after directly substitution. 0/0 is one of indeterminate form so we have to use another method to evaluate the limit since direct substitution does not work.

For a polynomial or fractional function, to evaluate a limit with another method if direct substitution does not work, you can do by using factorization method. Simply factor the expression of both denominator and numerator then cancel the same expression.

From x²-6x+5, you can factor as (x-5)(x-1) because -5-1 = -6 which is middle term and (-5)(-1) = 5 which is the last term.

From x²-25, you can factor as (x+5)(x-5) via differences of two squares.

After factoring the expressions, we get a new Limit.

\displaystyle \large{ \lim_{x\to 5}\frac{(x-5)(x-1)}{(x-5)(x+5)}}

We can cancel x-5.

\displaystyle \large{ \lim_{x\to 5}\frac{x-1}{x+5}}

Then directly substitute x = 5 in.

\displaystyle \large{ \lim_{x\to 5}\frac{5-1}{5+5}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{4}{10}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{2}{5}=\frac{2}{5}}

Therefore, the limit value is 2/5.

L’Hopital Method

I wouldn’t recommend using this method since it’s <em>too easy</em> but only if you know the differentiation. You can use this method with a limit that’s evaluated to indeterminate form. Most people use this method when the limit method is too long or hard such as Trigonometric limits or Transcendental function limits.

The method is basically to differentiate both denominator and numerator, do not confuse this with quotient rules.

So from the given function:

\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}

Differentiate numerator and denominator, apply power rules.

<u>Differential</u> (Power Rules)

\displaystyle \large{y = ax^n \longrightarrow y\prime= nax^{n-1}

<u>Differentiation</u> (Property of Addition/Subtraction)

\displaystyle \large{y = f(x)+g(x) \longrightarrow y\prime = f\prime (x) + g\prime (x)}

Hence from the expressions,

\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2-6x+5)}{\frac{d}{dx}(x^2-25)}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2)-\frac{d}{dx}(6x)+\frac{d}{dx}(5)}{\frac{d}{dx}(x^2)-\frac{d}{dx}(25)}}

<u>Differential</u> (Constant)

\displaystyle \large{y = c \longrightarrow y\prime = 0 \ \ \ \ \sf{(c\ \  is \ \ a \ \ constant.)}}

Therefore,

\displaystyle \large{ \lim_{x \to 5} \frac{2x-6}{2x}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{2(x-3)}{2x}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{x-3}{x}}

Now we can substitute x = 5 in.

\displaystyle \large{ \lim_{x \to 5} \frac{5-3}{5}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{2}{5}}=\frac{2}{5}

Thus, the limit value is 2/5 same as the first method.

Notes:

  • If you still get an indeterminate form 0/0 as example after using l’hopital rules, you have to differentiate until you don’t get indeterminate form.
8 0
2 years ago
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