Let $ a,\ b$ be real numbers satisfying $ \int_0^1 (ax\plus{}b)^2dx\equal{}1$.
Determine the values of $ a,\ b$ for which $ \int_0^1 3x(ax\plus{}b)\ dx$ is maximized.
1. **Set up the integral to be maximized:**
   We need to maximize the integral 
   \[
   f(a, b) = \int_{0}^{1} 3x(ax + b) \, dx.
   \]
   First, compute the integral:
   \[
   f(a, b) = \int_{0}^{1} 3x(ax + b) \, dx = \int_{0}^{1} 3ax^2 + 3bx \, dx.
   \]
   Evaluating the integral, we get:
   \[
   f(a, b) = 3a \int_{0}^{1} x^2 \, dx + 3b \int_{0}^{1} x \, dx = 3a \left[ \frac{x^3}{3} \right]_{0}^{1} + 3b \left[ \frac{x^2}{2} \right]_{0}^{1}.
   \]
   Simplifying, we find:
   \[
   f(a, b) = 3a \cdot \frac{1}{3} + 3b \cdot \frac{1}{2} = a + \frac{3b}{2}.
   \]

2. **Set up the constraint:**
   The constraint is given by:
   \[
   \int_{0}^{1} (ax + b)^2 \, dx = 1.
   \]
   Compute the integral:
   \[
   \int_{0}^{1} (ax + b)^2 \, dx = \int_{0}^{1} (a^2x^2 + 2abx + b^2) \, dx.
   \]
   Evaluating the integral, we get:
   \[
   \int_{0}^{1} a^2x^2 \, dx + \int_{0}^{1} 2abx \, dx + \int_{0}^{1} b^2 \, dx = a^2 \left[ \frac{x^3}{3} \right]_{0}^{1} + 2ab \left[ \frac{x^2}{2} \right]_{0}^{1} + b^2 \left[ x \right]_{0}^{1}.
   \]
   Simplifying, we find:
   \[
   \frac{a^2}{3} + ab + b^2 = 1.
   \]
   Multiplying through by 3, we get:
   \[
   a^2 + 3ab + 3b^2 = 3.
   \]

3. **Maximize the function subject to the constraint:**
   We need to maximize \( a + \frac{3b}{2} \) subject to \( a^2 + 3ab + 3b^2 = 3 \).

4. **Eliminate \( b \) from the equations:**
   Solve for \( b \) in terms of \( a \) and \( k \):
   \[
   a + \frac{3b}{2} = k \implies b = \frac{2k - 2a}{3}.
   \]
   Substitute \( b \) into the constraint:
   \[
   a^2 + 3a \left( \frac{2k - 2a}{3} \right) + 3 \left( \frac{2k - 2a}{3} \right)^2 = 3.
   \]
   Simplify the equation:
   \[
   a^2 + 2a(k - a) + \frac{4(k - a)^2}{3} = 3.
   \]
   Further simplification gives:
   \[
   a^2 + 2ak - 2a^2 + \frac{4k^2 - 8ak + 4a^2}{3} = 3.
   \]
   Combine like terms:
   \[
   3a^2 - 2a^2 + 2ak - 2ak + \frac{4k^2}{3} - \frac{8ak}{3} + \frac{4a^2}{3} = 3.
   \]
   Simplify further:
   \[
   a^2 - \frac{4a^2}{3} + \frac{4k^2}{3} - \frac{8ak}{3} = 3.
   \]
   Combine terms:
   \[
   \frac{3a^2 - 4a^2 + 4k^2 - 8ak}{3} = 3.
   \]
   Simplify:
   \[
   -a^2 + \frac{4k^2 - 8ak}{3} = 3.
   \]
   Multiply through by 3:
   \[
   -3a^2 + 4k^2 - 8ak = 9.
   \]
   Rearrange:
   \[
   4k^2 - 8ak - 3a^2 = 9.
   \]

5. **Solve the quadratic equation:**
   The quadratic equation in \( a \) is:
   \[
   3a^2 - 8ak + 4k^2 - 9 = 0.
   \]
   The discriminant \( \Delta \) of this quadratic equation must be non-negative for real solutions:
   \[
   \Delta = (8k)^2 - 4 \cdot 3 \cdot (4k^2 - 9) = 64k^2 - 48(4k^2 - 9).
   \]
   Simplify the discriminant:
   \[
   \Delta = 64k^2 - 192k^2 + 432 = -128k^2 + 432.
   \]
   For real solutions, we need:
   \[
   -128k^2 + 432 \geq 0 \implies 432 \geq 128k^2 \implies k^2 \leq \frac{432}{128} = \frac{27}{8}.
   \]
   Therefore:
   \[
   -\sqrt{\frac{27}{8}} \leq k \leq \sqrt{\frac{27}{8}}.
   \]
   Simplify the range for \( k \):
   \[
   -\sqrt{3.375} \leq k \leq \sqrt{3.375}.
   \]
   Since \( \sqrt{3.375} \approx 1.837 \), we have:
   \[
   -1.837 \leq k \leq 1.837.
   \]

6. **Find the maximum value of \( k \):**
   The maximum value of \( k \) is \( \sqrt{3} \approx 1.732 \).

7. **Determine the corresponding values of \( a \) and \( b \):**
   When \( k = \sqrt{3} \), the quadratic equation has a double root:
   \[
   3a^2 - 8a\sqrt{3} + 4(\sqrt{3})^2 - 9 = 0 \implies 3a^2 - 8a\sqrt{3} + 12 - 9 = 0 \implies 3a^2 - 8a\sqrt{3} + 3 = 0.
   \]
   Solving for \( a \):
   \[
   a = \sqrt{3}.
   \]
   Substitute \( a = \sqrt{3} \) into the constraint:
   \[
   \sqrt{3} + \frac{3b}{2} = \sqrt{3} \implies b = 0.
   \]

The final answer is \( \boxed{ a = \sqrt{3}, b = 0 } \).