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Kaylis [27]
3 years ago
9

F(x)=8x sqrt(x-x^2) Find the exact maximum

Mathematics
1 answer:
Mariana [72]3 years ago
3 0

First take note of the domain of <em>f(x)</em> ; the square root term is defined as long as <em>x</em> - <em>x</em> ² ≥ 0, or 0 ≤ <em>x</em> ≤ 1.

Check the value of <em>f(x)</em> at these endpoints:

<em>f</em> (0) = 0

<em>f</em> (1) = 0

Take the derivative of <em>f(x)</em> :

f(x)=8x\sqrt{x-x^2}=8x\left(x-x^2\right)^{\frac12}

\implies f'(x)=8\left(x-x^2\right)^{\frac12}+4x\left(x-x^2\right)^{-\frac12}(1-2x)=4\left(x-x^2\right)^{-\frac12}\left(2\left(x-x^2)\right)+x(1-2x)\right)=\dfrac{4(3x-4x^2)}{\sqrt{x-x^2}}

For <em>x</em> ≠ 0, we can eliminate the √<em>x</em> term in the denominator:

x\neq0\implies f'(x)=\dfrac{4\sqrt x (3-4x)}{\sqrt{1-x}}

<em>f(x)</em> has critical points where <em>f '(x)</em> is zero or undefined. We know about the undefined case, which occurs at the boundary of the domain of <em>f(x)</em>. Check where <em>f '(x)</em> = 0 :

√<em>x</em> (3 - 4<em>x</em>) = 0

√<em>x</em> = 0   <u>or</u>   3 - 4<em>x</em> = 0

The first case gives <em>x</em> = 0, which we ignore. The second leaves us with <em>x</em> = 3/4, at which point we get a maximum of max{<em>f(x) </em>} = 3√3 / 2.

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Step-by-step explanation:

Using Pythagoras theorem, I will solve all of the problems.

<h3>________________________________________________</h3>

<u>Question 9:</u>

  • 10² = 6² + x²
  • => 100 = 36 + x²
  • => 100 - 36 = x²
  • => 64 = x²
  • => x = 8
<h3>________________________________________________</h3>

<u>Question 10:</u>

  • 26² = 24² + x²
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  • => 676 - 576 = x²
  • => 100 = x²
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<h3>________________________________________________</h3>

<u>Question 11:</u>

  • 15² = 12² + x²
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  • => 225 - 144 = x²
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  • => x = 9
<h3>________________________________________________</h3>

<u>Question 12:</u>

  • x² = 8² + 12²
  • => x² = 64 + 144
  • => x² = 208
  • => x = √208
  • => x = 14.2 (Rounded)
<h3>________________________________________________</h3>

<u>Question 13:</u>

  • 7² = 2² + x²
  • => 49 = 4 + x²
  • => 49 - 4 = x²
  • => 45 = x²
  • => x = √45
  • => x = 6.7 (Rounded)
<h3>________________________________________________</h3>

<u>Question 14</u>

First, let's solve for the variable x using Pythagoras theorem.

  • => 5² = 3² + x²
  • => 25 = 9 + x²
  • => 16 = x²
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Now, let's solve for the variable y using Pythagoras theorem.

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Answers (Nearest tenth):

  • x = 4 units
  • y = 7.5 units
<h3>________________________________________________</h3>

<u>Question 15:</u>

First, let's find the value of the variable y using Pythagoras theorem.

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Now, let's find the value of the variable x using multiplication.

  • x = 2y
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Answer (Nearest tenth)

  • x = 10.6 units
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<h3>________________________________________________</h3>
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