/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 A triangle has one vertex at \((... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 2

A triangle has one vertex at \((0,0)\) and the other two on the graph of $y=-2 x^{2}+54\( at \)(x, y)\( and \)(-\mathrm{x}, \mathrm{y})$ where \(0<\mathrm{x}<\sqrt{27}\). The value of \(\mathrm{x}\) so that the corresponding triangle has maximum area is (A) \(\frac{\sqrt{27}}{2}\) (B) 3 (C) \(2 \sqrt{3}\) (D) None

Short Answer

Expert verified
Answer: 3.

Step by step solution

01

1. Analyze the vertices of the triangle in terms of x

The vertex at \((0,0)\) is the origin. The other two vertices have coordinates \((x,y)\) and \((-x,y)\), where \(y=-2x^{2}+54\).
02

2. Find the base and height of the triangle

The base of the triangle is the distance between the points \((x,y)\) and \((-x,y)\). Using the distance formula, we can find the length of the base in terms of x: \(2x\). Since the triangle is symmetrical with respect to the y-axis, its height (the distance from the origin to the parabola) is equal to the y-coordinate of the points on the parabola: \(y=-2x^2 + 54\)
03

3. Write an expression for the area of the triangle

The area of a triangle is given by the formula: Area \(=\frac{1}{2} {\cdot}\) Base \({\cdot}\) Height In our case, the base is \(2x\) and the height is \(y = -2x^2 + 54\). Therefore, the area of the triangle can be expressed as: Area \(=\frac{1}{2} {\cdot} 2x {\cdot} (-2x^2 + 54)\) Simplifying this expression, we get: Area \(=x(-2x^2 + 54)\)
04

4. Find the value of x that maximizes the area

To find the value of x that maximizes the area, we need to take the derivative of the Area function and find the critical points. The derivative is: \(\frac{d}{dx}(x(-2x^2 + 54)) = -6x^2 + 54\) Now, we set the derivative equal to zero to find the critical points: \(-6x^2 + 54 = 0\) After solving this quadratic equation, we get \(x = \sqrt{9} = 3\). Since \(0 < x < \sqrt{27}\), it satisfies our given range.
05

5. Compare the result that maximizes the area to the given options

We found the value of \(x\) that maximizes the area to be \(3\). Comparing this value to the given options, we see that it matches option (B). Therefore, the value of \(x\) that corresponds to the triangle with the maximum area is \(\boxed{3}\).

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