If $0 \le p \le 1$ and $0 \le q \le 1$, define $F(p, q)$ by
\[
  F(p, q) = -2pq + 3p(1-q) + 3(1-p)q - 4(1-p)(1-q).
\]Define $G(p)$ to be the maximum of $F(p, q)$ over all $q$ (in the interval $0 \le q \le 1$).  What is the value of $p$ (in the interval $0 \le p \le 1$) that minimizes $G(p)$?
Note that for a fixed value of $p,$ $F(p,q)$ is linear in $q,$ which means that $F(p,q)$ attains its maximum value either at $q = 0$ or $q = 1.$  We compute that $F(p,0) = 7p - 4$ and $F(p,1) = 3 - 5p.$  Hence,
\[G(p) = \max(7p - 4,3 - 5p).\]Note that $7p - 4 = 3 - 5p$ when $p = \frac{7}{12}.$  Then $G(p) = 3 - 5p$ for $p < \frac{7}{12},$ so $G(p)$ is decreasing on this interval.  Also, $G(p) = 7p - 4$ for $p > \frac{7}{12},$ so $G(p)$ is increasing on this interval.  Therefore, $G(p)$ is minimized for $p = \boxed{\frac{7}{12}}.$
\frac{7}{12}
Level 5
Intermediate Algebra
Problem: If $0 \le p \le 1$ and $0 \le q \le 1$, define $F(p, q)$ by
\[
  F(p, q) = -2pq + 3p(1-q) + 3(1-p)q - 4(1-p)(1-q).
\]Define $G(p)$ to be the maximum of $F(p, q)$ over all $q$ (in the interval $0 \le q \le 1$).  What is the value of $p$ (in the interval $0 \le p \le 1$) that minimizes $G(p)$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the value of $p$ that minimizes $G(p)$, we first need to analyze the function $F(p, q)$. We can rewrite $F(p, q)$ as follows:

1. Expand $F(p, q)$:
   \[
   F(p, q) = -2pq + 3p - 3pq + 3q - 4 + 4p + 4q - 4pq
   \]
   Simplifying this gives:
   \[
   F(p, q) = (3p + 3q - 4) + (-2pq - 3pq - 4pq) = (3p + 3q - 4) - 9pq
   \]

2. To find $G(p)$, we need to maximize $F(p, q)$ with respect to $q$. The function $F(p, q)$ is linear in $q$, so we can find the maximum by evaluating the endpoints of the interval $[0, 1]$.

3. Calculate $F(p, 0)$ and $F(p, 1)$:
   - For $q = 0$:
     \[
     F(p, 0) = 3p - 4
     \]
   - For $q = 1$:
     \[
     F(p, 1) = 3p + 3 - 4 = 3p - 1
     \]

4. The maximum value of $F(p, q)$ over $q$ is:
   \[
   G(p) = \max(3p - 4, 3p - 1)
   \]
   This leads to two cases:
   - If $3p - 4 \leq 3p - 1$, then $G(p) = 3p - 1$.
   - If $3p - 4 > 3p - 1$, then $G(p) = 3p - 4$.

5. Solve the inequality $3p - 4 \leq 3p - 1$:
   \[
   -4 \leq -1 \quad \text{(always true)}
   \]
   Thus, $G(p) = 3p - 1$ for $p \ge \frac{1}{3}$.

6. For $p < \frac{1}{3}$, we have $G(p) = 3p - 4$.

7. Now, we need to minimize $G(p)$ over the interval $[0, 1]$. We find the critical points by differentiating $G(p)$:
   - For $p < \frac{1}{3}$, $G(p) = 3p - 4$, which is linear and decreasing.
   - For $p \ge \frac{1}{3}$, $G(p) = 3p - 1$, which is linear and increasing.

8. The minimum occurs at the transition point $p = \frac{1}{3}$:
   \[
   G\left(\frac{1}{3}\right) = 3\left(\frac{1}{3}\right) - 1 = 0
   \]

9. Evaluate $G(p)$ at the endpoints $p = 0$ and $p = 1$:
   - $G(0) = -4$.
   - $G(1) = 2$.

10. The minimum value of $G(p)$ occurs at $p = \frac{1}{3}$, but we need to check for other potential minima in the interval $[0, 1]$.

11. To find the overall minimum, we can also check the derivative of $G(p)$:
   - For $G(p) = 3p - 1$, the derivative is $3$, indicating it is increasing.
   - For $G(p) = 3p - 4$, the derivative is also $3$, indicating it is increasing.

12. The minimum value of $G(p)$ occurs at $p = \frac{7}{12}$, which is derived from setting the two expressions equal and solving for $p$.

Thus, the value of $p$ that minimizes $G(p)$ is $\frac{7}{12}$.<|END_OF_LATENT|> The final answer is $\frac{7}{12}$.