Question

If x+y=20 and p=xy, what is the maximum value of p.



Well, i found the answer as 100.Rechecking with ur answer ❤❣️​

Answer 1

ANSWER :-

Maximum value of xy = 9

Proof,

Calculus Method:

Given, x+y = 6. Transposing x to RHS,

y = 6 - x

Substituting for y=6 - x,

xy = x(6-x) = 6x - x² = f(x), suppose……………………………………………..(1)

To find the maximum value of xy, we have to find the value of x for which f'(x) = 0 and then check if f(x) is maximum at such a value from the sign of f"(x). If f"(x) is negative (‹ 0) for a particular value of x, it means f(x) is maximum at that value of x.

Differentiating eq.(1) with respect to x,

f'(x) = df/dx = 6–2x

f"(x) = d²y/dx² = -2…………………………………………………………….(2)

Now f'(x) = 0 if 6–2x = 0

i.e. if 2x = 6

i.e. if x = 3

Since, f"(x) is negative (eq.(2)), xy is a maximum and the maximum value of xy is obtained by putting x = 3 in eq.(1).

Maximum of xy = 6x - x² = 6.3 - 3² = 18 - 9 = 9 (Proved)

(note :-if the upper answer is correct than like otherwise report. don't thank unusually. thank u)

Answer 2

$\large\underline{\sf{Solution-}}$

➢Given that

$\rm :\longmapsto\:x + y = 20$

$\rm :\implies\:y = 20 - x - - - (1)$

and

$\rm :\longmapsto\:p = xy - - - (2)$

➢Now, we have to find maximum value of p.

➢So, we use the concept of Maxima and Minima of differentiation.

➢Substituting the value of equation (1) in (2), we get

$\rm :\longmapsto\:p = x(20 - x)$

$\rm :\longmapsto\:p = 20x - {x}^{2}$

➢On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}p =\dfrac{d}{dx}( 20x - {x}^{2} )$

$\rm :\longmapsto\:\dfrac{dp}{dx} =\dfrac{d}{dx}20x - \dfrac{d}{dx}{x}^{2}$

$\rm :\longmapsto\:\dfrac{dp}{dx} =20 - 2x$

➢For maxima or minima,

$\rm :\longmapsto\:\dfrac{dp}{dx} =0$

$\rm :\longmapsto\:20 - 2x = 0$

$\rm :\longmapsto\:20 = 2x$

$\bf\implies \:x = 10 - - - (3)$

Now,

$\rm :\longmapsto\:\dfrac{d^{2} p}{dx ^{2} } =- 2$

$\rm :\longmapsto\:\dfrac{d^{2} p}{dx ^{2} } < 0$

$\bf\implies \:p \: is \: maximum \: when \: x = 10$

➢Substituting the value of x in equation (1), we get

$\rm :\longmapsto\:y = 20 - 10$

$\bf\implies \:y = 10$

Hence,

$\purple{\underbrace{\boxed{\sf{Maximum \: value \: of \: p = xy = 10 \times 10 = 100}}}}$

Basic Concept Used :-

HOW TO FIND MAXIMUM AND MINIMUM VALUE OF A FUNCTION

Let y = f(x) be the given function.

Differentiate the given function with respect to x.

let f'(x) = 0 and find critical points.

Then find the second derivative, say f''(x).

Apply the critical points in the second derivative to check the sign.

The function f (x) is maximum when f''(x) < 0.

The function f (x) is minimum when f''(x) > 0.