1dy/dx = 2(a-x)[1+2x/2√x2+b2] + 2(x+√x2+b2).-1
For maxima,dy/dx=0
from here we get:
a-x=√x2+b2
Squaring,a2-2ax+x2= x2+b2
Solving,x=(a2-b2)/2a
Thus maximum value of y= a2+b2
(substituting value of x to get y)
If x is real, the maximum value of -
y = 2\left(a-x\right) \{ x+ \sqrt{x^2 + b^2} \}
is ________ ?
-
UP 0 DOWN 0 0 7
7 Answers
62Avik.. I am a bit lazy
but see the derivative of x+√x2+b2
That will be very closely related to x+√x2+b2 [1]
So you will get an easy expression relatively..
Try to use that..
otherwise i will get out of my slumber..
13Yes sir, using calculus i got it [1],....
But the Qn was mentioned in the Algebra section (Quadratic)....so i was wondering how it cud be solved tht way....
1
66We have
y=2(a-x)(x+\sqrt{x^2+b^2})=\dfrac{2b^2(a-x)}{\sqrt{x^2+b^2}-x}
\Rightarrow\ y\sqrt{x^2+b^2}=xy+2b^2(a-x)=2ab^2+xy-2b^2x
Squaring the two sides, we get
y^2x^2+y^2b^2=4a^2b^4 + x^2y^2 + 4b^4x^2+4ab^2xy -4b^2x^2y -8ab^4x
which gives
y^2=4a^2b^2 + 4b^2x^2+4axy -4x^2y -8ab^2x
\Rightarrow\ 4x^2(b^2-y)+4ax(y-2b^2)+4a^2b^2-y^2=0
Since x is real, the discriminant must be non-negative. So
16a^2(y-2b^2)^2-16(b^2-y)(4a^2b^2-y^2)\geq 0
\Rightarrow\ y^2(a^2+b^2-y)\geq 0
As such we get
y\leq a^2+b^2
Hence, the maximum value of y is
\boxed{y_\mathrm{max}= a^2+b^2}
The equality occurs when
x_m=\dfrac{a(2b^2-y_\mathrm{max})}{2(b^2-y_\mathrm{max})}=\dfrac{a(b^2-a^2)}{-2a^2}=\dfrac{a^2-b^2}{2a}
24How abt this ?
Put t=x+√x2+b2
=> 1t=√x2+b2-xb2
=>t-b2t=2x
=> y=(2a-t+b2t)t=b2+2at-t2=a2+b2-(a-t)2
=>ymax=a2+b2 when x=(a2-b2)/2a