Math Problem Statement

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

Solution

To find the exact coordinates of the maximum point MM of the curve described by the equation

y=(x+5)32x,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 yy with respect to xx, i.e., dydx\frac{dy}{dx}, equals zero. Let's first find dydx\frac{dy}{dx}.

The given equation is: y=(x+5)32x.y = (x + 5)\sqrt{3} - 2x.

Taking the derivative with respect to xx: dydx=ddx((x+5)32x).\frac{dy}{dx} = \frac{d}{dx} \left( (x + 5)\sqrt{3} - 2x \right).

This simplifies to: dydx=32.\frac{dy}{dx} = \sqrt{3} - 2.

2. Solve dydx=0\frac{dy}{dx} = 0:

To find the stationary point, we set dydx=0\frac{dy}{dx} = 0: 32=0.\sqrt{3} - 2 = 0.

However, this expression is incorrect since 32\sqrt{3} - 2 is a constant that does not equal zero. This means that the curve does not have a stationary point at xx, 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.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Stationary Points
Maxima and Minima

Formulas

Derivative dy/dx
Solving dy/dx = 0

Theorems

First Derivative Test

Suitable Grade Level

Grades 11-12