Optimization: The Derivative Test

Can you solve each equation separately? If so, then this is a simple system to solve!

The first equation gives \(-{2}={2}x\text{,}\) so \(x=-1\text{.}\) The second one is equivalent to \({8}={16}y\text{,}\) so \(y={{\textstyle\frac{1}{2}}}\text{.}\)

Try adding or subtracting the two equations to find an equation with only one variable.

If we add these two equations together, the 3 and the \({6}y\) cancel giving \({16}x=0\text{,}\) so \(x=0\text{.}\) Setting \(x=0\) in either of the two equations gives \(y={-{\textstyle\frac{1}{2}}}\text{.}\) Thus, the only solution is \(x=0\) and \(y={-{\textstyle\frac{1}{2}}}\text{.}\)

Any pair of values \((x,y)\) which is a solution to the system must satisfy both equations. In particular, the \(y\)-coordinates have to equal at the same time \(x^2+{5}x-{2}\) and \(x+3\text{.}\) Use this to find the only possible values for \(x\) for which \((x,y)\) could be a solution.

Any pair of values \((x,y)\) which is a solution must satisfy both equations. In particular, the \(y\)-coordinates of any solution have to both equal \(x^2+{5}x-{2}\) and \(x+3\text{.}\) Thus, we have

which has solutions \(x=-{5}\) and \(x={1}\text{.}\) These are the only values of \(x\) for which \((x,y)\) can satisfy both equations in the system of equations. Setting \(x=-{5}\) in either of the two equations we find \(y=-{2}\text{,}\) so \((x,y)=(-{5},-{2})\) is a solution (the only one with \(x=-{5}\)). Setting \(x={1}\) in either of the two equations we find \(y={4}\text{,}\) so \((x,y)=({1},{4})\) is the only other solution.

Thus, the only solutions are \((x,y)=(-{5},-{2})\) and \((x,y)=({1},{4})\text{.}\)

Note that you can isolate \(y\) in both equations. Note that any pair of values \((x,y)\) which is a solution to the system must satisfy both equations. Use this to find the only possible values for \(x\) for which \((x,y)\) could be a solution.

Solving for \(y\) in both equations we find that

Any pair of values \((x,y)\) which is a solution must satisfy both of these equations. In particular, the \(y\)-coordinates of any solution have to both equal \(x^2-{24}\) and \(x-{12}\text{.}\)

Thus, we must have

which has solutions \(x=-{3}\) and \(x={4}\text{.}\) These are the only values of \(x\) for which \((x,y)\) can satisfy both equations in the system of equations. Setting \(x=-{3}\) in either of the two equations we find \(y=-{15}\text{,}\) so \((x,y)=(-{3},-{15})\) is a solution (the only one with \(x=-{3}\)). Setting \(x={4}\) in either of the two equations we find \(y=-{8}\text{,}\) so \((x,y)=({4},-{8})\) is the only other solution.

Note that you can solve for \(y\) in the second equation. Since any pair of values \((x,y)\) which is a solution to the system must satisfy both equations, then the \(y\)-value of any solution must be what the second equation tells you it should be. Plug this into the first equation to find the only possible values for \(x\) for which \((x,y)\) is a solution.

Solving for \(y\) in the second equation gives \(y=-x^2/{4}\text{.}\) Any pair of values \((x,y)\) which is a solution to the system must satisfy \(y=-x^2/{4}\text{,}\) and so we can plug this into the first equation, giving

so \(x=0\) or \(x=-1/{4}\text{.}\) These are the only two possible values of \(x\) for which \((x,y)\) may be a solution of the system (for some \(y\)). Setting \(x=0\) in the second equation of the system we find \(y=0\text{,}\) so \((x,y)=(0,0)\) is a solution (the only one with \(x=0\)). Setting \(x=-1/{4}\) in the second equation of the system we find \(y=-1/{64}\text{,}\) so \((x,y)=(-1/{4},-1/{64})\) is the only other solution.

A solution to the system is a pair of values \((x,y)\) which makes both equations true. Note that you can factor out a \(y\) in the second equation. What does the factored form tell you about the solutions to the second equation?

Factoring out a \({5}y\) in the second equation gives \({5}y(1+x)=0\text{,}\) so either \({5}y=0\) or \(1+x=0\) (or both). This shows that any solution \((x,y)\) to the system must have \(y=0\) or \(x=-1\text{.}\) Setting \(y=0\) in the first equation gives \[7x+3x^2=0,\] which has solutions \(x=0\) or \(x=-{{\textstyle\frac{7}{3}}}\text{.}\) Thus, \((x,y)=(0,0)\) and \((x,y)=(-{{\textstyle\frac{7}{3}}},0)\) are two solutions (the only solutions with \(y=0\)). Setting \(x=-1\) in the first equation gives \(-{4}+y^2=0\text{,}\) so \(y=\pm 2\text{.}\) Thus, \((x,y)=(-1,-2)\) and \((x,y)=(-1,2)\) are two more solutions (the only solutions with \(x=-1\)).

The solutions to the system are \((0,0)\text{,}\) \((-{{\textstyle\frac{7}{3}}},0)\text{,}\) \((-1,-2)\text{,}\) and \((-1,2)\text{.}\)

When finding a partial derivative all other variables are treated as constants.

When we differentiate with respect to \(x\text{,}\) we think of \(y\) as a constant. Thus,

When finding a partial derivative all other variables are treated as constants.

When we differentiate with respect to \(y\text{,}\) we think of \(x\) as a constant. Thus,

What is the relationship between the components of the gradient vector and the partial derivatives of the function?

We have \(\nabla f(x,y)=f_x(x,y)\vec{i}+f_y(x,y)\vec{j}\text{,}\) so

What is the relationship between the components of the gradient vector and the partial derivatives of the function?

We have \(\nabla g(x,y)=g_x(x,y)\vec{i}+g_y(x,y)\vec{j}\text{,}\) so

The determinant of

is \(ad-bc\text{.}\)

The determinant of

is \(ad-bc\text{.}\)