Answer:
[tex]\frac{140}{3}, \frac{140}{3}, \frac{140}{3}[/tex]
Step-by-step explanation:
Let the numbers be [tex]x, y\ and\ z.[/tex]
Such that:
[tex]x + y + z = 140[/tex]
Make z the subject
[tex]z = 140 -x - y[/tex]
For their product to be maximum, we have:
[tex]f(x,y,z) = xyz[/tex]
Substitute [tex]z = 140 -x - y[/tex] in [tex]f(x,y,z) = xyz[/tex]
[tex]f(x,y) = xy(140 - x - y)[/tex]
Open bracket
[tex]f(x,y) = 140xy - x^2y - xy^2[/tex]
Differentiate w.r.t x and y
[tex]f_x=140y - 2xy - y^2[/tex]
[tex]f_y=140x - x^2 - 2xy[/tex]
Since the products are maximum, then [tex]f_x = f_y = 0[/tex]
For [tex]f_x=140y - 2xy - y^2[/tex]
[tex]140y - 2xy - y^2 = 0[/tex]
Factorize:
[tex]y(140 - 2x - y) = 0[/tex]
Split
[tex]y = 0\ or\ 140 - 2x - y = 0[/tex]
Make y the subject
[tex]y = 0\ or\ y = 140 - 2x[/tex]
For [tex]f_y=140x - x^2 - 2xy[/tex]
[tex]140x - x^2 - 2xy = 0[/tex]
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Substitute y = 0
[tex]140x - x^2 -2x*0 = 0[/tex]
[tex]140x - x^2 = 0[/tex]
Factorize
[tex]x(140 - x)= 0[/tex]
[tex]x = 0\ or\ 140-x = 0[/tex]
[tex]x = 0\ or\ x = 140[/tex]
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Substitute [tex]y = 140 - 2x[/tex]
[tex]140x - x^2 - 2xy = 0[/tex]
[tex]140x - x^2 - 2x(140 - 2x) = 0[/tex]
[tex]140x - x^2 - 280x + 4x^2 = 0[/tex]
Re-arrange
[tex]4x^2 -x^2 +140x - 280x = 0[/tex]
[tex]3x^2 -140x = 0[/tex]
Factor x out
[tex]x(3x - 140) = 0[/tex]
Divide through by x
[tex]3x - 140 = 0[/tex]
[tex]3x = 140[/tex]
[tex]x = \frac{140}{3}[/tex]
Recall that: [tex]y = 140 - 2x[/tex]
[tex]y = 140 - 2 * \frac{140}{3}[/tex]
[tex]y = 140 - \frac{280}{3}[/tex]
Take LCM
[tex]y = \frac{140*3-280}{3}[/tex]
[tex]y = \frac{140}{3}[/tex]
Recall that:
[tex]z = 140 -x - y[/tex]
[tex]z = 140 - \frac{140}{3} - \frac{140}{3}[/tex]
Take LCM
[tex]z = \frac{3 * 140- 140 - 140}{3}[/tex]
[tex]z = \frac{140}{3}[/tex]
Hence, the numbers are:
[tex]\frac{140}{3}, \frac{140}{3}, \frac{140}{3}[/tex]
