CRE 4 - Maximization / Minimization

We know, AM ≥ GM [equality occurs when the numbers are equal]

∴ For two numbers, x and y

⇒ $\frac{x+y}{2}$ ≥ $\sqrt{xy}$

⇒ x + y ≥ 2 × √16

⇒ x + y ≥ 8

∴ Least possible value of (x + y) is 8. [This happen when x = y = 4.]

Note: When product of two numbers is constant, their sum is least possible when the two numbers are equal (or as close to each other as possible).

Hence, option (c).

We know, AM ≥ GM [equality occurs when the numbers are equal]

∴ For two numbers, x and y

⇒ $\frac{x+y}{2}$ ≥ $\sqrt{xy}$

⇒ 16 ≥ 2 × $\sqrt{xy}$

⇒ $\sqrt{xy}$ ≤ 8

⇒ xy ≤ 64

∴ Highest possible value of x × y is 64. [This happen when x = y = 8.]

Note: When sum of two numbers is constant, their product is highest possible when the two numbers are equal (or as close to each other as possible).

Hence, option (b).

In such cases we divided the coefficient of the variable with corresponding power and write the sum in following way.

x + y = $\frac{x}{2}$ + $\frac{x}{2}$ + y

Now we have three terms, $\frac{x}{2}$, $\frac{x}{2}$ and y.

We know, AM ≥ GM [equality occurs when the numbers are equal]

⇒ $\frac{{\displaystyle \frac{x}{2}}+\frac{x}{2}+y}{3}$ ≥ $\sqrt[3]{\frac{x}{2}\times \frac{x}{2}\times y}$

⇒ x + y ≥ 3 × $\sqrt[3]{\frac{x^{2}y}{4}}$

⇒ x + y ≥ 3 × $\sqrt[3]{\frac{32}{4}}$

⇒ x + y ≥ 3 × 2

⇒ x + y ≥ 6

∴ Least possible value of (x + y) is 6. [This happen when $\frac{x}{2}$ = $\frac{x}{2}$ = y = 2.]

Hence, option (b).

In such cases we divided the coefficient of the variable with corresponding power and write the sum in following way.

2x +3y = x + x + y + y + y

Now we have five terms, x, x, y, y and y.

We know, AM ≥ GM [equality occurs when the numbers are equal]

⇒ $\frac{x+x+y+y+y}{5}$ ≥ $\sqrt[5]{x\times x\times y\times y\times y}$

⇒ 2x + 3y ≥ 5 × $\sqrt[5]{x^2\times y^3}$

⇒ 2x + 3y ≥ 5 × $\sqrt[5]{32}$

⇒ 2x + 3y ≥ 10

∴ Least possible value of (2x + 3y) is 10. [This happen when x = x = y = y = y = 2]

Hence, option (c).

In such cases we divided the coefficient of the variable with corresponding power and write the sum in following way.

3x + 2y = $\frac{3x}{2}$ + $\frac{3x}{2}$ + $\frac{2y}{3}$ + $\frac{2y}{3}$ + $\frac{2y}{3}$

Now we have five terms, $\frac{3x}{2}$, $\frac{3x}{2}$, $\frac{2y}{3}$, $\frac{2y}{3}$ and $\frac{2y}{3}$.

We know, AM ≥ GM [equality occurs when the numbers are equal]

⇒ $\frac{{\displaystyle \frac{3x}{2}}+\frac{3x}{2}+{\displaystyle \frac{2y}{3}}+\frac{2y}{3}+\frac{2y}{3}}{5}$ ≥ $\sqrt[5]{\frac{3x}{2}\times \frac{3x}{2}\times \frac{2y}{3}\times \frac{2y}{3}\times \frac{2y}{3}}$

⇒ 2x + 3y ≥ 5 × $\sqrt[5]{\frac{2}{3}{x}^2\times y^3}$

⇒ 2x + 3y ≥ 5 × $\sqrt[5]{\frac{2}{3}\times 48}$

⇒ 2x + 3y ≥ 10

∴ Least possible value of (2x + 3y) is 10. [This happen when $\frac{3x}{2}$ = $\frac{3x}{2}$ = $\frac{2y}{3}$ = $\frac{2y}{3}$ = $\frac{2y}{3}$ = 2]

Hence, option (c).

In such cases we divided the coefficient of the variable with corresponding power and write the sum in following way.

3x + y = $\frac{3x}{2}$ + $\frac{3x}{2}$ + y

Now we have three terms, $\frac{3x}{2}$, $\frac{3x}{2}$, and y.

We know, AM ≥ GM [equality occurs when the numbers are equal]

⇒ $\frac{{\displaystyle \frac{3x}{2}}+\frac{3x}{2}+y}{3}$ ≥ $\sqrt[3]{\frac{3x}{2}\times \frac{3x}{2}\times y}$

⇒ $\frac{3x+y}{3}$ ≥ $\sqrt[3]{\frac{9x^{2}y}{4}}$

⇒ $\frac{9}{3}$ ≥ $\sqrt[3]{\frac{9x^{2}y}{4}}$

⇒ 3 ≥ $\sqrt[3]{\frac{9x^{2}y}{4}}$

⇒ $\frac{9x^{2}y}{4}$ ≤ 27

⇒ x²y ≤ 12

∴ Highest possible value of (3x + y) is 12. [This happen when $\frac{3x}{2}$ = $\frac{3x}{2}$ = y = 3]

Hence, option (b).

Let the three natural numbers be a, b and c

⇒ a + b + c = 35

We need to find the maximum value of their product.

We know, when sum of two or more numbers is constant, their product is highest possible when the numbers are equal (or as close to each other as possible).

Since, the three numbers are distinct, to maximize their product the numbers should be as close to each other as possible.

Hence, we need to write 35 as a sum of 3 distinct integers which are as to each other as possible.

∴ 35 = 10 + 12 + 13

∴ a × b × c = 10 × 12 × 13 = 1560

Hence, 1560.

Given, x + y + z = 6.

Let us consider three numbers (x + y), (y + z) and (z + x)

We know, AM ≥ GM

⇒ $\frac{(x+y)+(y+z)+(z+x)}{3}$ ≥ $\sqrt[3]{(x + y)(y + z)(z + x)}$

⇒ $\frac{2(x+y+z)}{3}$ ≥ $\sqrt[3]{(x + y)(y + z)(z + x)}$

⇒ $\frac{2\times 6}{3}$ ≥ $\sqrt[3]{(x + y)(y + z)(z + x)}$

⇒ 4 ≥ $\sqrt[3]{(x + y)(y + z)(z + x)}$

⇒ (x + y)(y + z)(z + x) ≤ 64

Hence, 64.