CAT 2025 Lesson : Inequalities - Sum or Product is Constant

If $2x + 3y = 60$, what is the maximum value of $xy$?

Solution

As $2x + 3y = 60$, $2x \times 3y = 6xy$ will be maximum when $\bm{2x = 3y}$.
(Note: When 6$xy$ is maximised, this maximises $xy$ as well, as 6 is a constant.)

⇒ 2$x$ $=$ 3$y$ $=$ $\dfrac{60}{2}$ $=$ $30$

⇒ $x$ $=$ $15$ and $y$ $=$ $10$

Maximum value of $xy$ $=$ $15 \times 10$ $=$ 150

Answer: $150$

Where $x$ is a positive number, what is the minimum value of $3 x + \dfrac{27}{x}$?

Solution

Product of the 2 terms is a constant ⇒ $3 x \times \dfrac{27}{x} = 81$

∴ Sum of the 2 terms is minimum when they are equal ⇒ $3 x = \dfrac{27}{x}$ ⇒ $x^2 = 9$ ⇒ $x = 3$

Minimum value of $3 x + \dfrac{27}{x}$ $=$ $3 \times 3 + \dfrac{27}{3} = 18$

Answer: $18$

If $x + y + z = 20$, then the maximum value of $(x + 3)(y - 1)(z + 2)$ is

Solution

Sum of the terms $= (x + 3) + (y - 1) + (z + 2)$ = $x + y + z + 4$ = $\bm{24}$, which is a constant

Product of the terms will be maximum when the three terms are equal.
i.e., $(x + 3) = (y - 1) = (z + 2) = \dfrac{24}{3} = 8$

$\therefore$ Maximum value of $(x + 3) (y - 1) (z + 2) = 8 \times 8 \times 8 = 512$

Answer: $512$

$p, q$ and $r$ are three non-negative integers such that $p + q + r = 10$. The maximum value of $pq + qr + pr + pqr$ is
[XAT 2013]

(1) $\ge 40$ and $< 50 \space \space \space$
(2) $\ge 50$ and $< 60\space \space \space$
(3) $\ge 60$ and $< 70\space \space \space$
(4) $\ge 70$ and $< 80\space \space \space$
(5) $\ge 80$ and $< 90\space \space \space\space \space \space$

Solution

When sum of terms is constant, the their product will be maximum when they are equal or close to each other. In other words, the difference between them should be minimal.

$p, q$ and $r$ have to be integers. Therefore, the closest values they can take are $3, 3$ and $4$.

Maximum value of $pq + qr + pr + pqr = 3^{2} + (3 \times 4) + (3 \times 4) + (3^{2} \times 4)$

$= 9 + 12 + 12 + 36 = \bm{69}$

Answer: (3) $\ge 60$ and $< 70$

Sikander has to construct a rectangle with a perimeter of $50$ cm, such that $l^{3}b^{2}$ is maximum (where $l$ and $b$ are the length and breadth of the rectangle respectively). What is the difference between the length and the breadth?

Solution

Perimeter of rectangle $= 2(l + b) = 50$
⇒ $l + b = 25$ -----(1)

$l^{3}b^{2}$ is maximum when $\dfrac{l}{3} = \dfrac{b}{2}$⇒ $l = \dfrac{3b}{2}$

Substituting into (1), we get $l = 15$ and $b = 10$

$\therefore l - b = 5$

Answer: $5$