It is given that $f(x)=ax^3+bx^2+cx+d$, where $a$, $b$, $c$ and $d$ are constants. The curve $C$ with equation $y=f(x)$ passes through $(0,-1)$ and has a maximum point at $(-1,1)$. The area bounded by $C$, the $x$-axis and the lines $x=2$ and $x=3$ is $\frac{31}{4}$ units$^2$. Given that $f(x)>0$ for $2<x<3$, find the values of $a$, $b$, $c$ and $d.

1. Find the value of $d$ using the given $y$-intercept.

The curve passes through $(0, -1)$.
$f(0) = a(0)^3 + b(0)^2 + c(0) + d = -1$
$\implies d = -1$

2. Form an equation using the coordinates of the maximum point.

The curve passes through $(-1, 1)$.
$f(-1) = a(-1)^3 + b(-1)^2 + c(-1) - 1 = 1$
$-a + b - c = 2 \quad \text{--- (1)}$

3. Form a second equation using the property of a stationary point.

Since $(-1, 1)$ is a maximum point, $f'(-1) = 0$.
$f'(x) = 3ax^2 + 2bx + c$
$f'(-1) = 3a(-1)^2 + 2b(-1) + c = 0$
$3a - 2b + c = 0 \quad \text{--- (2)}$

4. Form a third equation using the given area under the curve.

Since $f(x) > 0$ for $2 < x < 3$, the area is given by the definite integral:
$\int_2^3 (ax^3 + bx^2 + cx - 1) \, dx = \frac{31}{4}$
$\left[ \frac{a}{4}x^4 + \frac{b}{3}x^3 + \frac{c}{2}x^2 - x \right]_2^3 = \frac{31}{4}$
$\left( \frac{81}{4}a + 9b + \frac{9}{2}c - 3 \right) - \left( 4a + \frac{8}{3}b + 2c - 2 \right) = \frac{31}{4}$
$\frac{65}{4}a + \frac{19}{3}b + \frac{5}{2}c - 1 = \frac{31}{4}$
$\frac{65}{4}a + \frac{19}{3}b + \frac{5}{2}c = \frac{35}{4}$
Multiplying by 12 gives:
$195a + 76b + 30c = 105 \quad \text{--- (3)}$

5. Solve the system of linear equations to find $a, b$, and $c$.

From (1) and (2):
$(1) + (2): 2a - b = 2 \implies b = 2a - 2$
Substitute $b$ into (2):
$3a - 2(2a - 2) + c = 0 \implies c = a - 4$
Substitute $b$ and $c$ into (3):
$195a + 76(2a - 2) + 30(a - 4) = 105$
$195a + 152a - 152 + 30a - 120 = 105$
$377a - 272 = 105 \implies 377a = 377 \implies a = 1$
Then $b = 2(1) - 2 = 0$ and $c = 1 - 4 = -3$.

(Alternatively, a graphing calculator can be used to solve the linear system formed by (1), (2), and (3) directly.)