a) Find the maximum value of f(x,y) = x^{3}y^{4} for x,y \geq 0 on the unit circle.
Answers
The maximum value of f(x,y) on the unit of circle is [tex]\frac{48 \sqrt{21}}{2,401}[/tex].
The full question is in the attachment. From the unit circle
x² + y² = 1y² = 1 - x²Substitute to f(x,y)
f(x,y) = x³y⁴ = x³ (y²)² = x³ (1 - x²)²
f(x,y) = x³ (1 - 2x² + x⁴)
f(x,y) = x³ - 2x⁵ + x⁷
df/dx = 3x² - (2 × 5)x⁴ + 7x⁶
df/dx = 3x² - 10x⁴ + 7x⁶
df/dx = x² (3 - 10x² + 7x⁴)
To find the maximum or minimum value, the derivated equals zero.
0 = x² (7x⁴ - 10x² + 3)
0 = x² (x - 1) (x + 1) (7x² - 3)
x² = 0x = 0
f(x,y) = x³ - 2x⁵ + x⁷ = 0³ - (2 × 0⁵) + 0⁷ = 0x - 1 = 0
x = 1
f(x,y) = 1³ - (2 × 1⁵) + 1⁷ = 1 - 2 + 1 = 0x + 1 = 0
x = - 1
f(x,y) = (- 1)³ - (2 × (- 1)⁵) + (- 1)⁷ = - 1 + 2 - 1 = 0
7x² - 3 = 0
Using ABC formula
a = 7b = 0c = - 3[tex]x_{1,2} \:=\: \frac{-b \pm \sqrt{b^2 \:-\: 4ac}}{2a}[/tex]
[tex]x_{1,2} \:=\: \frac{-0 \pm \sqrt{0^2 \:-\: (4 \times 7 \times - 3)}}{2 \times 7}[/tex]
[tex]x_{1,2} \:=\: \frac{\pm \sqrt{4 \times 21)}}{14}[/tex]
[tex]x_{1,2} \:=\: \frac{\pm 2 \sqrt{21}}{14}[/tex]
[tex]x_{1,2} \:=\: \frac{\pm \sqrt{21}}{7}[/tex]
[tex]x_1 \:=\: \frac{\pm \sqrt{21}}{7}[/tex]f(x,y) = x³ - 2x⁵ + x⁷
[tex]f \:=\: (\frac{\sqrt{21}}{7})^3 \:-\: (2 \times (\frac{\sqrt{21}}{7})^5) \:+\: (\frac{\sqrt{21}}{7})^7[/tex]
[tex]f \:=\: \frac{21 \sqrt{21}}{7^3} \:-\: \frac{2 \times 21 \times 21 \sqrt{21}}{7^5} \:+\: \frac{21 \times 21 \times 21 \sqrt{21}}{7^7}[/tex]
[tex]f \:=\: \frac{7^4 \times 21 \sqrt{21} \:-\: 7^2 \times 2 \times 21 \times 21 \sqrt{21} \:+\: 21 \times 21 \times 21 \sqrt{21}}{7^7}[/tex]
[tex]f \:=\: \frac{7^5 \times 3 \sqrt{21} \:-\: 7^4 \times 2 \times 3 ^2 \sqrt{21} \:+\: 7^3 \times 3^3 \sqrt{21}}{7^7}[/tex]
[tex]f \:=\: \frac{7^3 \: (7^2 \times 3 \sqrt{21} \:-\: 7 \times 2 \times 3 ^2 \sqrt{21} \:+\: 3^3 \sqrt{21})}{7^7}[/tex]
[tex]f \:=\: \frac{147 \sqrt{21} \:-\: 126 \sqrt{21} \:+\: 27 \sqrt{21}}{7^4}[/tex]
[tex]f \:=\: \frac{48 \sqrt{21}}{2,401}[/tex][tex]x_2 \:=\: - \frac{\sqrt{21}}{7}[/tex]
[tex]f \:=\: (- \frac{\sqrt{21}}{7})^3 \:-\: (2 \times (- \frac{\sqrt{21}}{7})^5) \:+\: (- \frac{\sqrt{21}}{7})^7[/tex]
[tex]f \:=\: \frac{- 147 \sqrt{21} \:+\: 126 \sqrt{21} \:-\: 27 \sqrt{21}}{7^4}[/tex]
[tex]f \:=\: \frac{- 48 \sqrt{21}}{2,401}[/tex]
The maximum value from x₁.
The minimum value from x₂.
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Related Questions
Use the points form of a line to write the equation of a line thag that has a slope of 2 and passing through the points (-4,3) write the equation in slope intercept form
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If the equation is passing trough the origin, it will be passing trough the point (0,0). We now for our problem that the equations is also passing trough the point (-4,3). So, our line is passing trough the points (0,0) and (-4,3). To write the equation in slope-intercept form, first, we need to find its slope . To do that we are going to use the slope formula: .
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B)
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C)
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Answer:
A
Step-by-step explanation
The circle is filled in which means that -10 is included. and it is going to the right which means it is greater than -10. Hope this helps!
PLEASE HELP WILL MARK BRAINLIEST
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Answers
Answer:
Area of rectangle possible. : 6ft² ; 10 ft², 12 ft²
Step-by-step explanation:
Feets of fencing = 14
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Perimeter = 2(x + y)
For x = 1
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Area of rectangle ; x *y = 1 * 6 = 6 ft²
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Step-by-step explanation:
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Answer:
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Step-by-step explanation:
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