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Sagot :
Certainly! Let's work through finding the critical points, domain endpoints, and local extreme values for the function:
[tex]\[ y = 5x \sqrt{64 - x^2} \][/tex]
Step 1: Determine the domain.
The expression inside the square root, [tex]\( 64 - x^2 \)[/tex], must be non-negative for the function to be real:
[tex]\[ 64 - x^2 \geq 0 \][/tex]
[tex]\[ x^2 \leq 64 \][/tex]
[tex]\[ -8 \leq x \leq 8 \][/tex]
So, the domain is [tex]\([-8, 8]\)[/tex].
Step 2: Find the critical points.
To find the critical points, we need to take the derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex] and set it equal to zero.
[tex]\[ y = 5x \sqrt{64 - x^2} \][/tex]
Using the product rule and the chain rule, the derivative [tex]\( \frac{dy}{dx} \)[/tex] is found to be:
[tex]\[ \frac{dy}{dx} = 5 \sqrt{64 - x^2} + 5x \cdot \frac{-x}{\sqrt{64 - x^2}} \][/tex]
Simplifying:
[tex]\[ \frac{dy}{dx} = 5 \sqrt{64 - x^2} - \frac{5x^2}{\sqrt{64 - x^2}} \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{5(64 - x^2) - 5x^2}{\sqrt{64 - x^2}} \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{320 - 10x^2}{\sqrt{64 - x^2}} \][/tex]
Setting [tex]\( \frac{dy}{dx} = 0 \)[/tex]:
[tex]\[ 320 - 10x^2 = 0 \][/tex]
[tex]\[ 10x^2 = 320 \][/tex]
[tex]\[ x^2 = 32 \][/tex]
[tex]\[ x = \pm 4\sqrt{2} \][/tex]
So, the critical points are [tex]\( x = -4\sqrt{2} \)[/tex] and [tex]\( x = 4\sqrt{2} \)[/tex].
Step 3: Evaluate the function at the critical points and endpoints.
We now evaluate the original function [tex]\( y \)[/tex] at the critical points and the endpoints of the domain [tex]\( [-8, 8] \)[/tex].
For [tex]\( x = -4\sqrt{2} \)[/tex]:
[tex]\[ y(-4\sqrt{2}) = 5 \cdot (-4\sqrt{2}) \cdot \sqrt{64 - (-4\sqrt{2})^2} \][/tex]
[tex]\[ y(-4\sqrt{2}) = 5 \cdot (-4\sqrt{2}) \cdot \sqrt{64 - 32} \][/tex]
[tex]\[ y(-4\sqrt{2}) = 5 \cdot (-4\sqrt{2}) \cdot \sqrt{32} \][/tex]
[tex]\[ y(-4\sqrt{2}) = 5 \cdot (-4\sqrt{2}) \cdot 4\sqrt{2} \][/tex]
[tex]\[ y(-4\sqrt{2}) = 5 \cdot (-4\sqrt{2}) \cdot 4\sqrt{2} = -5 \cdot 16 \cdot 2 = -160 \][/tex]
For [tex]\( x = 4\sqrt{2} \)[/tex]:
[tex]\[ y(4\sqrt{2}) = 5 \cdot (4\sqrt{2}) \cdot \sqrt{64 - (4\sqrt{2})^2} \][/tex]
[tex]\[ y(4\sqrt{2}) = 5 \cdot (4\sqrt{2}) \cdot \sqrt{64 - 32} \][/tex]
[tex]\[ y(4\sqrt{2}) = 5 \cdot (4\sqrt{2}) \cdot \sqrt{32} \][/tex]
[tex]\[ y(4\sqrt{2}) = 5 \cdot (4\sqrt{2}) \cdot 4\sqrt{2} \][/tex]
[tex]\[ y(4\sqrt{2}) = 5 \cdot 16 \cdot 2 = 160 \][/tex]
For [tex]\( x = -8 \)[/tex]:
[tex]\[ y(-8) = 5 \cdot (-8) \cdot \sqrt{64 - (-8)^2} \][/tex]
[tex]\[ y(-8) = 5 \cdot (-8) \cdot \sqrt{64 - 64} \][/tex]
[tex]\[ y(-8) = 5 \cdot (-8) \cdot \sqrt{0} = 0 \][/tex]
For [tex]\( x = 8 \)[/tex]:
[tex]\[ y(8) = 5 \cdot 8 \cdot \sqrt{64 - 8^2} \][/tex]
[tex]\[ y(8) = 5 \cdot 8 \cdot \sqrt{64 - 64} \][/tex]
[tex]\[ y(8) = 5 \cdot 8 \cdot \sqrt{0} = 0 \][/tex]
Step 4: Identify local extrema.
From the evaluations, we have:
[tex]\[ (-4\sqrt{2}, -160), \quad (4\sqrt{2}, 160), \quad (-8, 0), \quad (8, 0) \][/tex]
Comparing the function values, we find that the local maximum value is [tex]\( 160 \)[/tex] at [tex]\( x = 4\sqrt{2} \)[/tex]. Therefore, the local maxima occurs at the point [tex]\( (4\sqrt{2}, 160) \)[/tex].
Conclusion:
From the critical points and domain endpoints, the point corresponding to the local maxima is:
[tex]\[ \boxed{(4\sqrt{2}, 160)} \][/tex]
Choosing the correct option:
A. The point(s) corresponding to the local maxima is/are [tex]\(\boxed{(4\sqrt{2}, 160)}\)[/tex]
Option B is incorrect since there are points corresponding to the local maxima.
[tex]\[ y = 5x \sqrt{64 - x^2} \][/tex]
Step 1: Determine the domain.
The expression inside the square root, [tex]\( 64 - x^2 \)[/tex], must be non-negative for the function to be real:
[tex]\[ 64 - x^2 \geq 0 \][/tex]
[tex]\[ x^2 \leq 64 \][/tex]
[tex]\[ -8 \leq x \leq 8 \][/tex]
So, the domain is [tex]\([-8, 8]\)[/tex].
Step 2: Find the critical points.
To find the critical points, we need to take the derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex] and set it equal to zero.
[tex]\[ y = 5x \sqrt{64 - x^2} \][/tex]
Using the product rule and the chain rule, the derivative [tex]\( \frac{dy}{dx} \)[/tex] is found to be:
[tex]\[ \frac{dy}{dx} = 5 \sqrt{64 - x^2} + 5x \cdot \frac{-x}{\sqrt{64 - x^2}} \][/tex]
Simplifying:
[tex]\[ \frac{dy}{dx} = 5 \sqrt{64 - x^2} - \frac{5x^2}{\sqrt{64 - x^2}} \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{5(64 - x^2) - 5x^2}{\sqrt{64 - x^2}} \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{320 - 10x^2}{\sqrt{64 - x^2}} \][/tex]
Setting [tex]\( \frac{dy}{dx} = 0 \)[/tex]:
[tex]\[ 320 - 10x^2 = 0 \][/tex]
[tex]\[ 10x^2 = 320 \][/tex]
[tex]\[ x^2 = 32 \][/tex]
[tex]\[ x = \pm 4\sqrt{2} \][/tex]
So, the critical points are [tex]\( x = -4\sqrt{2} \)[/tex] and [tex]\( x = 4\sqrt{2} \)[/tex].
Step 3: Evaluate the function at the critical points and endpoints.
We now evaluate the original function [tex]\( y \)[/tex] at the critical points and the endpoints of the domain [tex]\( [-8, 8] \)[/tex].
For [tex]\( x = -4\sqrt{2} \)[/tex]:
[tex]\[ y(-4\sqrt{2}) = 5 \cdot (-4\sqrt{2}) \cdot \sqrt{64 - (-4\sqrt{2})^2} \][/tex]
[tex]\[ y(-4\sqrt{2}) = 5 \cdot (-4\sqrt{2}) \cdot \sqrt{64 - 32} \][/tex]
[tex]\[ y(-4\sqrt{2}) = 5 \cdot (-4\sqrt{2}) \cdot \sqrt{32} \][/tex]
[tex]\[ y(-4\sqrt{2}) = 5 \cdot (-4\sqrt{2}) \cdot 4\sqrt{2} \][/tex]
[tex]\[ y(-4\sqrt{2}) = 5 \cdot (-4\sqrt{2}) \cdot 4\sqrt{2} = -5 \cdot 16 \cdot 2 = -160 \][/tex]
For [tex]\( x = 4\sqrt{2} \)[/tex]:
[tex]\[ y(4\sqrt{2}) = 5 \cdot (4\sqrt{2}) \cdot \sqrt{64 - (4\sqrt{2})^2} \][/tex]
[tex]\[ y(4\sqrt{2}) = 5 \cdot (4\sqrt{2}) \cdot \sqrt{64 - 32} \][/tex]
[tex]\[ y(4\sqrt{2}) = 5 \cdot (4\sqrt{2}) \cdot \sqrt{32} \][/tex]
[tex]\[ y(4\sqrt{2}) = 5 \cdot (4\sqrt{2}) \cdot 4\sqrt{2} \][/tex]
[tex]\[ y(4\sqrt{2}) = 5 \cdot 16 \cdot 2 = 160 \][/tex]
For [tex]\( x = -8 \)[/tex]:
[tex]\[ y(-8) = 5 \cdot (-8) \cdot \sqrt{64 - (-8)^2} \][/tex]
[tex]\[ y(-8) = 5 \cdot (-8) \cdot \sqrt{64 - 64} \][/tex]
[tex]\[ y(-8) = 5 \cdot (-8) \cdot \sqrt{0} = 0 \][/tex]
For [tex]\( x = 8 \)[/tex]:
[tex]\[ y(8) = 5 \cdot 8 \cdot \sqrt{64 - 8^2} \][/tex]
[tex]\[ y(8) = 5 \cdot 8 \cdot \sqrt{64 - 64} \][/tex]
[tex]\[ y(8) = 5 \cdot 8 \cdot \sqrt{0} = 0 \][/tex]
Step 4: Identify local extrema.
From the evaluations, we have:
[tex]\[ (-4\sqrt{2}, -160), \quad (4\sqrt{2}, 160), \quad (-8, 0), \quad (8, 0) \][/tex]
Comparing the function values, we find that the local maximum value is [tex]\( 160 \)[/tex] at [tex]\( x = 4\sqrt{2} \)[/tex]. Therefore, the local maxima occurs at the point [tex]\( (4\sqrt{2}, 160) \)[/tex].
Conclusion:
From the critical points and domain endpoints, the point corresponding to the local maxima is:
[tex]\[ \boxed{(4\sqrt{2}, 160)} \][/tex]
Choosing the correct option:
A. The point(s) corresponding to the local maxima is/are [tex]\(\boxed{(4\sqrt{2}, 160)}\)[/tex]
Option B is incorrect since there are points corresponding to the local maxima.
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