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Digiron [165]
3 years ago
5

Find three positive numbers whose sum is 140 and whose product is a maximum. (Enter your answers as a comma-separated list.)

Mathematics
1 answer:
Paladinen [302]3 years ago
8 0

Answer:

\frac{140}{3}, \frac{140}{3}, \frac{140}{3}

Step-by-step explanation:

Let the numbers be x, y\ and\ z.

Such that:

x + y + z = 140

Make z the subject

z = 140 -x - y

For their product to be maximum, we have:

f(x,y,z) = xyz

Substitute z = 140 -x - y in f(x,y,z) = xyz

f(x,y) = xy(140 - x - y)

Open bracket

f(x,y) = 140xy - x^2y - xy^2

Differentiate w.r.t x and y

f_x=140y - 2xy - y^2

f_y=140x - x^2 - 2xy

Since the products are maximum, then f_x = f_y = 0

For f_x=140y - 2xy - y^2

140y - 2xy - y^2 = 0

Factorize:

y(140 - 2x - y) = 0

Split

y = 0\ or\ 140 - 2x - y = 0

Make y the subject

y = 0\ or\ y = 140 - 2x

For f_y=140x - x^2 - 2xy

140x - x^2 - 2xy = 0

---------------------------------------------------

Substitute y = 0

140x - x^2 -2x*0 = 0

140x - x^2 = 0

Factorize

x(140 - x)= 0

x = 0\ or\ 140-x = 0

x = 0\ or\ x = 140

---------------------------------------------------

Substitute y = 140 - 2x

140x - x^2 - 2xy = 0

140x - x^2 - 2x(140 - 2x) = 0

140x - x^2 - 280x + 4x^2 = 0

Re-arrange

4x^2 -x^2 +140x - 280x = 0

3x^2 -140x = 0

Factor x out

x(3x - 140) = 0

Divide through by x

3x - 140 = 0

3x = 140

x = \frac{140}{3}

Recall that: y = 140 - 2x

y = 140 - 2 * \frac{140}{3}

y = 140 - \frac{280}{3}

Take LCM

y = \frac{140*3-280}{3}

y = \frac{140}{3}

Recall that:

z = 140 -x - y

z = 140 - \frac{140}{3} - \frac{140}{3}

Take LCM

z =  \frac{3 * 140- 140 - 140}{3}

z =  \frac{140}{3}

Hence, the numbers are:

\frac{140}{3}, \frac{140}{3}, \frac{140}{3}

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From the options,

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