To find the largest possible value of \(\( x \)\) for the given problem, we start by applying the Pythagorean theorem to the cone. The theorem relates the slant height \(\( l \)\), the radius \(\( r \)\), and the height \(\( h \)\) of the cone through the equation:
\(\[ l^2 = r^2 + h^2 \]\)
Given:
- Radius (\(\( r \)\)) = \(\( x + 3 \)\)
- Height (\(\( h \)\)) = \(\( 5x - 3 \)\)
- Slant height (\(\( l \)\)) = \(\( \sqrt{15} \)\)
We substitute these expressions into the Pythagorean theorem:
\(\[ (\sqrt{15})^2 = (x + 3)^2 + (5x - 3)^2 \]\)
First, we simplify the left side of the equation:
\(\[ 15 = (x + 3)^2 + (5x - 3)^2 \]\)
Next, we expand both squared terms on the right side:
\(\[ (x + 3)^2 = x^2 + 6x + 9 \]\)
\(\[ (5x - 3)^2 = 25x^2 - 30x + 9 \]\)
Now, add these two expressions:
\(\[ 15 = x^2 + 6x + 9 + 25x^2 - 30x + 9 \]\)
\(\[ 15 = x^2 + 25x^2 + 6x - 30x + 9 + 9 \]\)
\(\[ 15 = 26x^2 - 24x + 18 \]\)
To solve for \(\( x \)\), we first move all terms to one side of the equation to set it to zero:
\(\[ 26x^2 - 24x + 18 - 15 = 0 \]\)
\(\[ 26x^2 - 24x + 3 = 0 \]\)
This is a quadratic equation in standard form \(\( ax^2 + bx + c = 0 \)\), where \(\( a = 26 \)\), \(\( b = -24 \)\), and \(\( c = 3 \)\).
We solve the quadratic equation using the quadratic formula:
\(\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]\)
Substituting \(\( a \)\), \(\( b \)\), and \(\( c \)\) into the formula:
\(\[ x = \frac{-(-24) \pm \sqrt{(-24)^2 - 4 \cdot 26 \cdot 3}}{2 \cdot 26} \]\)
\(\[ x = \frac{24 \pm \sqrt{576 - 312}}{52} \]\)
\(\[ x = \frac{24 \pm \sqrt{264}}{52} \]\)
\(\[ x = \frac{24 \pm 2\sqrt{66}}{52} \]\)
\(\[ x = \frac{12 \pm \sqrt{66}}{26} \]\)
Thus, the solutions for \(\( x \)\) are:
\(\[ x = \frac{12 - \sqrt{66}}{26} \]\)
\(\[ x = \frac{12 + \sqrt{66}}{26} \]\)
To find the largest possible value of \(\( x \)\), we select the larger of the two solutions:
\(\[ x = \frac{12 + \sqrt{66}}{26} \]\)
So, the largest possible value of \(\( x \)\) is \(\( \frac{12 + \sqrt{66}}{26} \)\).
