$x+y=10$

$max(x+y+xy+1)=?$

$=10+max(xy)+1$

$10+5\times 5+1=36$

($x\times y$ will be maximum when we take $x=y$)

$max(x+y+xy+1)=?$

$=10+max(xy)+1$

$10+5\times 5+1=36$

($x\times y$ will be maximum when we take $x=y$)

3 votes

It is given that $e^a+e^b=10$ where $a$ and $b$ are real. Then the maximum value of $(e^a+e^b+e^{a+b}+1)$ is

- $36$
- $\infty$
- $25$
- $21$

3 votes

$\underline{\mathbf{Answer:A}}$

$\underline{\mathbf{Solution:}}$

$\mathrm{e^a+e^b = 10,\;\;a \in R\;\;\;[Given]} \tag{1}$

Let $\mathrm{x = e^a}$, and $\mathrm {y = e^b}$

Then,

$\mathrm{\mathbf{max} (e^a+e^b+e^ae^b + 1) = \mathbf{max}(x+y+xy+1) = x+y+\mathbf{max}(xy) +1}$

Now, $\mathbf{xy}$ is maximum only when: $\mathrm{x = 5,\;y = 5\tag{2}}$

$\therefore \text{Maximum Value} = \underbrace{10}_\text{from (1)} + \underbrace{25}_\text{from(2)} + 1 = 36$

$\therefore \mathbf A$ is the correct option.

$\underline{\mathbf{Solution:}}$

$\mathrm{e^a+e^b = 10,\;\;a \in R\;\;\;[Given]} \tag{1}$

Let $\mathrm{x = e^a}$, and $\mathrm {y = e^b}$

Then,

$\mathrm{\mathbf{max} (e^a+e^b+e^ae^b + 1) = \mathbf{max}(x+y+xy+1) = x+y+\mathbf{max}(xy) +1}$

Now, $\mathbf{xy}$ is maximum only when: $\mathrm{x = 5,\;y = 5\tag{2}}$

$\therefore \text{Maximum Value} = \underbrace{10}_\text{from (1)} + \underbrace{25}_\text{from(2)} + 1 = 36$

$\therefore \mathbf A$ is the correct option.