# Using Calculus to find 2 Unknowns

• May 6th 2013, 04:03 AM
JellyOnion
Using Calculus to find 2 Unknowns
Little stuck on this question. Any help greatly appreciated.

Find two positive numbers whose sum is 4 and such that the sum of the cube of the first is and the square of the second is as small as possible.

Cheers
• May 6th 2013, 04:15 AM
Prove It
Re: Using Calculus to find 2 Unknowns
You should first write up your equations. Can you do that?
• May 6th 2013, 05:11 AM
JellyOnion
Re: Using Calculus to find 2 Unknowns
X+y=4
X^3+y^2=???

That's where I get stuck
• May 6th 2013, 05:31 AM
HallsofIvy
Re: Using Calculus to find 2 Unknowns
From X+ y= 4, y= 4- X so the function you want to minimize is X^3+ (4- X)^2= X^3+ X^2- 8X+ 16. Do you know what to do now?
• May 6th 2013, 05:34 AM
topsquark
Re: Using Calculus to find 2 Unknowns
Quote:

Originally Posted by JellyOnion
X+y=4
X^3+y^2=???

That's where I get stuck

As a typical example, say you have 30 m of fencing to surround a rectangular area and you need to find the maximum such area. You have the equation
P = 2x + 2y = 30
and you need to maximize the area: A = xy. So you solve the perimeter equation for y: y = 15 - x, then put that into the area formula: A = xy = x(15 - x). Then you take the derivative and set it to zero, etc.

So you've got x + y = 4 and you want to minimize $x^3 + y^2$. It's the same process.

-Dan
• May 6th 2013, 05:37 AM
Prove It
Re: Using Calculus to find 2 Unknowns
OK, you are wanting to minimise \displaystyle \begin{align*} f(x, y) = x^3 + y^2 \end{align*} subject to \displaystyle \begin{align*} g(x, y) = x + y = 4 \end{align*}. To minimise this we require using Lagrange Multipliers, i.e. to solve the equations

\displaystyle \begin{align*} \nabla f(x, y) &= \lambda \nabla g(x, y) \\ \left( 3x^2 , 2y \right) &= \lambda \left( 1, 1 \right) \\ \\ \begin{cases} 3x^2 &= \lambda \\ 2y &= \lambda \end{cases} \\ \\ 2y &= 3x^2 \\ y &= \frac{3}{2}x^2 \end{align*}

And since we know by our constraint that \displaystyle \begin{align*} x + y = 4 \end{align*}, that means

\displaystyle \begin{align*} x + \frac{3}{2}x^2 &= 4 \\ 3x^2 + 2x &= 8 \\ 3x^2 + 2x - 8 &= 0 \\ 3x^2 + 6x - 4x - 8 &= 0 \\ 3x \left( x + 2 \right) - 4 \left( x + 2 \right) &= 0 \\ \left( x + 2 \right) \left( 3x - 4 \right) &= 0 \\ x = -2 \textrm{ or } x = \frac{4}{3} \end{align*}

Since x and y have to be positive, that means \displaystyle \begin{align*} x = \frac{4}{3} \end{align*} and \displaystyle \begin{align*} y = \frac{8}{3} \end{align*}.

You can find the minimum value by subbing into the original function.
• May 7th 2013, 04:24 AM
JellyOnion
Re: Using Calculus to find 2 Unknowns
Cheers everyone