The problem describes a trapezoidal flower bed bounded by $y = x + 7$, the x-axis, and the vertical lines $x = 1$ and $x = a$, where $a > 1$. The area $A$ of the trapezoid is given by $A = \frac{1}{2}a^2 + 7a - \frac{15}{2}$. We need to find the value of $a$ when the area is 23 square meters.

1. Problem Description

The problem describes a trapezoidal flower bed bounded by $y = x + 7$, the x-axis, and the vertical lines $x = 1$ and $x = a$, where $a > 1$. The area $A$ of the trapezoid is given by $A = \frac{1}{2}a^2 + 7a - \frac{15}{2}$. We need to find the value of $a$ when the area is 23 square meters.

2. Solution Steps

We are given the area of the trapezoid as a function of $a$:

$A = \frac{1}{2}a^2 + 7a - \frac{15}{2}$

We are also given that the area is 23, so we set $A = 23$:

$23 = \frac{1}{2}a^2 + 7a - \frac{15}{2}$

Multiply both sides by 2 to eliminate fractions:

$46 = a^2 + 14a - 15$

Rearrange to form a quadratic equation:

$a^2 + 14a - 15 - 46 = 0$

$a^2 + 14a - 61 = 0$

Now we use the quadratic formula to solve for $a$:

$a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In our equation, $a^2 + 14a - 61 = 0$, we have $a = 1$, $b = 14$, and $c = -61$. Plugging these values into the quadratic formula:

$a = \frac{-14 \pm \sqrt{14^2 - 4(1)(-61)}}{2(1)}$

$a = \frac{-14 \pm \sqrt{196 + 244}}{2}$

$a = \frac{-14 \pm \sqrt{440}}{2}$

$a = \frac{-14 \pm \sqrt{4 \cdot 110}}{2}$

$a = \frac{-14 \pm 2\sqrt{110}}{2}$

$a = -7 \pm \sqrt{110}$

Since $a > 1$, we take the positive root:

$a = -7 + \sqrt{110}$

Since $\sqrt{100} = 10$ and $\sqrt{121} = 11$, $\sqrt{110}$ is between 10 and

1

1. So, $-7 + \sqrt{110}$ is approximately $10.488 - 7 = 3.488 > 1$.

$a = -7 - \sqrt{110}$ is a negative number, so we discard it because $a > 1$.

3. Final Answer

$a = -7 + \sqrt{110}$