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# Find three positive numbers whose sum is 100 and whose product is a maximum.

**Solution:**

Let the three numbers be x, y and z.

x + y + z = 100 -----(1)

⇒ z = 100 - x - y ------(2)

Product, P = x × y × z -----(3)

Substituting value of ‘z’ from (2) in (3),

P = x × y × (100 - x - y)

P = 100xy - x^{2}y - xy^{2}

⇒ f(x,y) = P = 100xy - x^{2}y - xy^{2}

δf/δx = 100y - 2xy - y^{2}

δf/δy = 100x - x^{2 }- 2xy

Consider δf/δx = 0

100y - 2xy - y^{2} = 0

y (100 - 2x - y) = 0

⇒y = 0 and (100 - 2x - y) = 0

y = 100 - 2x ------(4)

Consider δf/δy = 0

100x - x^{2}- 2xy = 0

x (100 - x - 2y) = 0------(5)

Substitute y = (100 - 2x) in (5)

⇒x [100 - x - 2(100 - 2x)] = 0

x (-100 + 3x) = 0

x = 0 or 3x - 100 = 0

⇒ x = 0 or **x =100 / 3**

Substitute x = 100 / 3 in (4)

y = 100 - 2x

y = 100 - 2(100 / 3)

**y = 100 / 3**

Let A = 𝑓_{xx} = -2x, B = 𝑓_{xy} = 100 - 2x -2y, C = 𝑓_{yy} = -2y

D = AC - B^{2}

D = [(-2x) (-2y)] - [100 - 2x - 2y]^{2}

D = [4 × 100/3 × 100/3] - [100 - 2(100/3) - 2(100/3)]^{2}

D = 40000/9 - 10000/9 = 30000/9

Now, D = 30000/9 > 0

And z = 100 - 100/3 -100/3

**z = 100/3**

∴ Maximum product is at (100/3, 100/3, 100/3).

## Find three positive numbers whose sum is 100 and whose product is a maximum.

**Summary:**

The three positive numbers whose sum is 100 and whose product is a maximum are (100/3, 100/3, 100/3).

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