Find maximum and minimum of $f(x,y)=x+y$

Question

I would appreciate if somebody could help me with the following problem:

Question: Find maximum and minimum of $f(x,y)=x+y$ when $$x^3+y^3=1,x,y \geq 0$$ I tried but couldn’t get it that way.

Answer 1

hint: use the Lagrange Multiplier function $f(x,y,\lambda)=x+y+\lambda(x^3+y^3-1)$ and compute the solutions of $f_{x}=0$ $f_{y}=0$ $f_{\lambda}=0$

Answer 2

This can be done using basic calculus as well.
We know that $$y=(1-x^3)^{\frac{1}{3}}$$
Let $$z=x+y=x+(1-x^3)^{\frac{1}{3}}$$
Then $$\frac{dz}{dx}=1-\frac{3x^2}{3(1-x^3)^{\frac{2}{3}}}$$
Putting $\frac{dz}{dx}=0$
$$x^2=(1-x^3)^{\frac{1}{3}}$$
We will get two answers on which we can perform double differentiation and find out if they are minima or maxima.
Once we get the value of $x$, we can plug it in and find out the value of $z$.

Answer 3

Parametrize the curve in terms of sine and cosine, use the inequalities to get the range of the argument ( theta ). Try using derivative tests.

Answer 4

We have $$ x^3+y^3 = 1 $$ $ x, y \geq 0$ If we solve for $y$ in terms of $x$ we obtain $$ y= (1-x^3)^{\frac{1}{3}} $$ and so we have to find the minimum of $$ f(x) = x+ (1-x^3)^{\frac{1}{3}} $$ The derivative of this function is $$ f'(x) = 1 - x^2(1-x^3)^{\frac{-2}{3}} $$ Solving for $f'(x) = 0$ we get $$ 1 = x^2(1-x^3)^{\frac{-2}{3}} $$ which after cubing both sides becomes $$ 1 = x^6 (1-x^3)^{-2} $$ or equivalently $$ (1-x^3)^2 = x^6 $$ so $$ 1-2x^3+x^6 = x^6 $$ which simplifies to $$ 1-2x^3 = 0 $$ $$ x^3 = \frac{1}{2}$$ $$ x = \left(\frac{1}{2}\right)^{\frac{1}{3}}$$ So the maximum or minimum of the function will occur at this point. Now, checking the boundary values, which are 0 and 1 since $x$ must be between 0 and 1 for both $x$ and $y$ to be positive and $x^3+y^3=1$, we find that at $x=0, y=1 $ and $x=1 , y=0$ the value of $f(x,y)$ is 1 and that at the point with the maximum or minimum, $x=y=\left(\frac{1}{2}\right)^{\frac{1}{3}}$ we have $$ f(x,y) = \left(\frac{1}{2}\right)^{\frac{1}{3}} + \left(\frac{1}{2}\right)^{\frac{1}{3}} = 2\left(\frac{1}{2}\right)^{\frac{1}{3}} = 2^{\frac{2}{3}}$$ Since this is greater than 1, this is the maximum of our function and the two minimums are at the boundaries, where the value is 1.