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Dmitrij [34]
4 years ago
10

Find two positive numbers satisfying the given requirements. the product is 192 and the sum of the first plus three times the se

cond is a minimum. (first number) (second number)
Mathematics
1 answer:
Zepler [3.9K]4 years ago
6 0
<h3>Given</h3>

Two positive numbers x and y such that xy = 192

<h3>Find</h3>

The values that minimize x + 3y

<h3>Solution</h3>

y = 192/x . . . . . solve for y

f(x) = x + 3y

f(x) = x + 3(192/x) . . . . . the function we want to minimize

We can find the x that minimizes of f(x) by setting the derivative of f(x) to zero.

... f'(x) = 1 - 576/x² = 0

... 576 = x² . . . . . . . . . . . . multiply by x², add 576

... √576 = x = 24 . . . . . . . take the square root

... y = 192/24 = 8 . . . . . . . find the value of y using the above equation for y

The first number is 24.

The second number is 8.

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Help please, I attached the question. Is it a!? <br> 
AVprozaik [17]

Answer:

A

Step-by-step explanation:

Recall that for a quadratic equation of the form:

0=ax^2+bx+c

The number of solutions it has can be determined using its discriminant:  

\Delta = b^2-4ac

Where:

  • If the discriminant is positive, we have two real solutions.
  • If the discriminant is negative, we have no real solutions.
  • And if the discriminant is zero, we have exactly one solution.

We have the equation:

2x^2+5x-k=0

Thus, <em>a</em> = 2, <em>b</em> = 5, and <em>c</em> = -<em>k</em>.

In order for the equation to have exactly one distinct solution, the discriminant must equal zero. Hence:

b^2-4ac=0

Substitute:

(5)^2-4(2)(-k)=0

Solve for <em>k</em>. Simplify:

25+8k=0

Solve:

\displaystyle k = -\frac{25}{8}

Thus, our answer is indeed A.

3 0
3 years ago
Read 2 more answers
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