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To find the exact coordinates of the maximum point \( M \) of the curve described by the equation

\[
y = (x + 5)\sqrt{3} - 2x,
\]

we need to follow these steps:

### 1. Differentiate the equation to find the stationary points:
The maximum occurs at a stationary point where the derivative of \( y \) with respect to \( x \), i.e., \( \frac{dy}{dx} \), equals zero. Let's first find \( \frac{dy}{dx} \).

The given equation is:
\[
y = (x + 5)\sqrt{3} - 2x.
\]

Taking the derivative with respect to \( x \):
\[
\frac{dy}{dx} = \frac{d}{dx} \left( (x + 5)\sqrt{3} - 2x \right).
\]

This simplifies to:
\[
\frac{dy}{dx} = \sqrt{3} - 2.
\]

### 2. Solve \( \frac{dy}{dx} = 0 \):
To find the stationary point, we set \( \frac{dy}{dx} = 0 \):
\[
\sqrt{3} - 2 = 0.
\]

However, this expression is incorrect since \( \sqrt{3} - 2 \) is a constant that does not equal zero. This means that the curve does not have a stationary point at \( x \), which would indicate that the maximum point might occur at the boundaries.

Upon reviewing, we can check the behavior at extreme points or boundaries of the domain. Would you like further exploration on that? 

Also, feel free to ask about any details, and here are some related questions:

1. How can we check for boundaries when solving for maximum points?
2. What is the role of differentiation in locating maxima and minima?
3. How can we calculate the second derivative to confirm concavity?
4. What is the interpretation of non-zero derivative constants in optimization?
5. Can we find maximum points by analyzing the endpoints of a curve?

**Tip:** Always check both the first and second derivatives when trying to locate and confirm maximum or minimum points on a curve.

Find the exact coordinates of M for the curve y = (x + 5)√3 − 2x, where M is the maximum point of the curve.

Solve the problem of finding the exact coordinates of the maximum point M on the curve y = (x + 5)√3 − 2x by using differentiation to find stationary points and applying the first derivative test. This step-by-step solution is suitable for students in Grades 11-12.

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