CUET UG 2025 — Mathematics Algebra

When the maximum value of $Z$Z occurs at two corner points, $Z$Z has the same value at both points.

At point $\left(\frac{112}{59}, \frac{135}{59}\right)$(59112​,59135​):

$$Z = p \cdot \frac{112}{59} + q \cdot \frac{135}{59}$$Z=p⋅59112​+q⋅59135​

$$Z = \frac{112p}{59} + \frac{135q}{59}$$Z=59112p​+59135q​

At point $(0, 3)$(0,3):

$$Z = p \cdot 0 + q \cdot 3$$Z=p⋅0+q⋅3

$$Z = 3q$$Z=3q

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Since both points give the maximum value:

$$\frac{112p}{59} + \frac{135q}{59} = 3q$$59112p​+59135q​=3q

$$\frac{112p}{59} = 3q - \frac{135q}{59}$$59112p​=3q−59135q​

$$\frac{112p}{59} = \frac{177q}{59} - \frac{135q}{59}$$59112p​=59177q​−59135q​

$$\frac{112p}{59} = \frac{42q}{59}$$59112p​=5942q​

$$112p = 42q$$112p=42q

$$8p = 3q$$8p=3q

Therefore, the relationship between $p$p and $q$q is $8p = 3q$8p=3q.