Question

Find three positive numbers whose sum is\(\;{\bf{100}}\)and whose product is a maximum.

Short answer

\(x = \frac{{100}}{3},y = \frac{{100}}{3},z = \frac{{100}}{3}\)

Step-by-step solution

Second derivative test

Suppose the second partial derivatives of\(f\)are continuous on a disk with center\({\bf{(a,b)}}\), and suppose that\({f_x}(a,b) = 0\)and\({f_y}(a,b) = 0\). Let

\(D = D(a,b) = {f_{xx}}(a,b){f_{yy}}(a,b) - {\left( {{f_{xy}}(a,b)} \right)^2}\)

(a) If\(D > 0\)and\({f_{xx}}(a,b) > 0\), then\(f(a,b)\)is a local minimum.

(b) If\(D > 0\)and\({f_{xx}}(a,b) < 0\), then\(f(a,b)\)is a local maximum.

(c) If\(D < 0\), then\(f(a,b)\)is not a local maximum or minimum.

Form equation

Let the three positive numbers be\(x,y,z\).

According to question:

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

The product \( = xyz\)

Using (1) \(z = 100 - x - y\)

Then the product\( = xy(100 - x - y)\)

\(\begin{aligned}{c}f(x,y) = xy(100 - x - y)\\ = 100xy - {x^2}y - x{y^2}\end{aligned}\)

Find critical point

The partial derivative of function is given by:

\(\begin{aligned}{l}{f_x} = 100y - 2xy - {y^2}\\{f_y} = 100x - {x^2} - 2xy\end{aligned}\)

Put\({f_x} = 0,{f_y} = 0\)

i.e. \(y(100 - 2x - y) = 0\)

And \(x(100 - x - 2y) = 0\)

On solving we find the critical point\(\left( {\frac{{100}}{3},\frac{{100}}{3}} \right)\).

Compute discriminant

Point\((0,0)\)is not critical point as it is not feasible since it is gives\(xyz = 0\)

Now \({f_{xx}} = - 2y\)

\(\begin{aligned}{l}{f_{yy}} = - 2x\\{f_{xy}} = 100 - 2x - 2y\end{aligned}\)

Then

\(\begin{aligned}{c}D = {f_{xx}}{f_{yy}} - f_{xy}^2\\ = 4xy - (100 - 2x - 2y)\end{aligned}\)

Compute discriminant at critical point

\(\begin{aligned}{c}{\rm{ At }}\left( {\frac{{100}}{3},\frac{{100}}{3}} \right);\\D = \frac{{40000}}{9} - {\left( {100 - \frac{{200}}{3} - \frac{{200}}{3}} \right)^2}\\ = \frac{{40000}}{9} - \frac{{10000}}{9}\\ = \frac{{10000}}{3} > 0\end{aligned}\)

And \({f_{xx}}\left( {\frac{{100}}{3},\frac{{100}}{3}} \right) = \frac{{ - 200}}{3} < 0\)

Find numbers

So \(\left( {\frac{{100}}{3},\frac{{100}}{3}} \right)\) is a point of maxima

i.e; \(x = \frac{{100}}{3},y = \frac{{100}}{3}\), "\(f\) " will have maximum value

So product of the three numbers\(x,y,z\)will be maximum

Then the required numbers are\(x = \frac{{100}}{3},y = \frac{{100}}{3},z = 100 - \frac{{100}}{3} - \frac{{100}}{3} = \frac{{100}}{3}\)

\(x = \frac{{100}}{3},y = \frac{{100}}{3},z = \frac{{100}}{3}\)