# 4.7 Maxima/minima problems  (Page 7/10)

 Page 7 / 10

## Key concepts

• A critical point of the function $f\left(x,y\right)$ is any point $\left({x}_{0},{y}_{0}\right)$ where either ${f}_{x}\left({x}_{0},{y}_{0}\right)={f}_{y}\left({x}_{0},{y}_{0}\right)=0,$ or at least one of ${f}_{x}\left({x}_{0},{y}_{0}\right)$ and ${f}_{y}\left({x}_{0},{y}_{0}\right)$ do not exist.
• A saddle point is a point $\left({x}_{0},{y}_{0}\right)$ where ${f}_{x}\left({x}_{0},{y}_{0}\right)={f}_{y}\left({x}_{0},{y}_{0}\right)=0,$ but $\left({x}_{0},{y}_{0}\right)$ is neither a maximum nor a minimum at that point.
• To find extrema of functions of two variables, first find the critical points, then calculate the discriminant and apply the second derivative test.

## Key equations

• Discriminant
$D={f}_{xx}\left({x}_{0},{y}_{0}\right){f}_{yy}\left({x}_{0},{y}_{0}\right)-{\left({f}_{xy}\left({x}_{0},{y}_{0}\right)\right)}^{2}$

For the following exercises, find all critical points.

$f\left(x,y\right)=1+{x}^{2}+{y}^{2}$

$f\left(x,y\right)={\left(3x-2\right)}^{2}+{\left(y-4\right)}^{2}$

$\left(\frac{2}{3},4\right)$

$f\left(x,y\right)={x}^{4}+{y}^{4}-16xy$

$f\left(x,y\right)=15{x}^{3}-3xy+15{y}^{3}$

$\left(0,0\right)$ $\left(\frac{1}{15},\frac{1}{15}\right)$

For the following exercises, find the critical points of the function by using algebraic techniques (completing the square) or by examining the form of the equation. Verify your results using the partial derivatives test.

$f\left(x,y\right)=\sqrt{{x}^{2}+{y}^{2}+1}$

$f\left(x,y\right)=\text{−}{x}^{2}-5{y}^{2}+8x-10y-13$

Maximum at $\left(4,-1,8\right)$

$f\left(x,y\right)={x}^{2}+{y}^{2}+2x-6y+6$

$f\left(x,y\right)=\sqrt{{x}^{2}+{y}^{2}}+1$

Relative minimum at $\left(0,0,1\right)$

For the following exercises, use the second derivative test to identify any critical points and determine whether each critical point is a maximum, minimum, saddle point, or none of these.

$f\left(x,y\right)=\text{−}{x}^{3}+4xy-2{y}^{2}+1$

$f\left(x,y\right)={x}^{2}{y}^{2}$

The second derivative test fails. Since ${x}^{2}{y}^{2}>0$ for all x and y different from zero, and ${x}^{2}{y}^{2}=0$ when either x or y equals zero (or both), then the absolute minimum occurs at $\left(0,0\right).$

$f\left(x,y\right)={x}^{2}-6x+{y}^{2}+4y-8$

$f\left(x,y\right)=2xy+3x+4y$

$f\left(-2,-\frac{3}{2}\right)=-6$ is a saddle point.

$f\left(x,y\right)=8xy\left(x+y\right)+7$

$f\left(x,y\right)={x}^{2}+4xy+{y}^{2}$

$f\left(0,0\right)=0;$ $\left(0,0,0\right)$ is a saddle point.

$f\left(x,y\right)={x}^{3}+{y}^{3}-300x-75y-3$

$f\left(x,y\right)=9-{x}^{4}{y}^{4}$

$f\left(0,0\right)=9$ is a local maximum.

$f\left(x,y\right)=7{x}^{2}y+9x{y}^{2}$

$f\left(x,y\right)=3{x}^{2}-2xy+{y}^{2}-8y$

Relative minimum located at $\left(2,6\right).$

$f\left(x,y\right)=3{x}^{2}+2xy+{y}^{2}$

$f\left(x,y\right)={y}^{2}+xy+3y+2x+3$

$\left(1,-2\right)$ is a saddle point.

$f\left(x,y\right)={x}^{2}+xy+{y}^{2}-3x$

$f\left(x,y\right)={x}^{2}+2{y}^{2}-{x}^{2}y$

$\left(2,1\right)$ and $\left(-2,1\right)$ are saddle points; $\left(0,0\right)$ is a relative minimum.

$f\left(x,y\right)={x}^{2}+y-{e}^{y}$

$f\left(x,y\right)={e}^{\text{−}\left({x}^{2}+{y}^{2}+2x\right)}$

$\left(-1,0\right)$ is a relative maximum.

$f\left(x,y\right)={x}^{2}+xy+{y}^{2}-x-y+1$

$f\left(x,y\right)={x}^{2}+10xy+{y}^{2}$

$\left(0,0\right)$ is a saddle point.

$f\left(x,y\right)=\text{−}{x}^{2}-5{y}^{2}+10x-30y-62$

$f\left(x,y\right)=120x+120y-xy-{x}^{2}-{y}^{2}$

The relative maximum is at $\left(40,40\right).$

$f\left(x,y\right)=2{x}^{2}+2xy+{y}^{2}+2x-3$

$f\left(x,y\right)={x}^{2}+x-3xy+{y}^{3}-5$

$\left(\frac{1}{4},\frac{1}{2}\right)$ is a saddle point and $\left(1,1\right)$ is the relative minimum.

$f\left(x,y\right)=2xy{e}^{\text{−}{x}^{2}-{y}^{2}}$

For the following exercises, determine the extreme values and the saddle points. Use a CAS to graph the function.

[T] $f\left(x,y\right)=y{e}^{x}-{e}^{y}$

A saddle point is located at $\left(0,0\right).$

[T] $f\left(x,y\right)=x\phantom{\rule{0.2em}{0ex}}\text{sin}\left(y\right)$

[T] $f\left(x,y\right)=\text{sin}\left(x\right)\text{sin}\left(y\right),x\in \left(0,2\pi \right),y\in \left(0,2\pi \right)$

There is a saddle point at $\left(\pi ,\pi \right),$ local maxima at $\left(\frac{\pi }{2},\frac{\pi }{2}\right)\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\left(\frac{3\pi }{2},\frac{3\pi }{2}\right),$ and local minima at $\left(\frac{\pi }{2},\frac{3\pi }{2}\right)\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\left(\frac{3\pi }{2},\frac{\pi }{2}\right).$

Find the absolute extrema of the given function on the indicated closed and bounded set $R.$

$f\left(x,y\right)=xy-x-3y;$ $R$ is the triangular region with vertices $\left(0,0\right),\left(0,4\right),\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\left(5,0\right).$

Find the absolute maximum and minimum values of $f\left(x,y\right)={x}^{2}+{y}^{2}-2y+1$ on the region $R=\left\{\left(x,y\right)|{x}^{2}+{y}^{2}\le 4\right\}.$

$\left(0,1,0\right)$ is the absolute minimum and $\left(0,-2,9\right)$ is the absolute maximum.

$f\left(x,y\right)={x}^{3}-3xy-{y}^{3}$ on $R=\left\{\left(x,y\right)\text{:}\phantom{\rule{0.2em}{0ex}}-2\le x\le 2,-2\le y\le 2\right\}$

$f\left(x,y\right)=\frac{-2y}{{x}^{2}+{y}^{2}+1}$ on $R=\left\{\left(x,y\right)\text{:}\phantom{\rule{0.2em}{0ex}}{x}^{2}+{y}^{2}\le 4\right\}$

There is an absolute minimum at $\left(0,1,-1\right)$ and an absolute maximum at $\left(0,-1,1\right).$

Find three positive numbers the sum of which is $27,$ such that the sum of their squares is as small as possible.

Find the points on the surface ${x}^{2}-yz=5$ that are closest to the origin.

$\left(\sqrt{5},0,0\right),\left(\text{−}\sqrt{5},0,0\right)$

Find the maximum volume of a rectangular box with three faces in the coordinate planes and a vertex in the first octant on the line $x+y+z=1.$

The sum of the length and the girth (perimeter of a cross-section) of a package carried by a delivery service cannot exceed $108$ in. Find the dimensions of the rectangular package of largest volume that can be sent.

$\text{18 by 36 by 18 in}.$

A cardboard box without a lid is to be made with a volume of $4$ ft 3 . Find the dimensions of the box that requires the least amount of cardboard.

Find the point on the surface $f\left(x,y\right)={x}^{2}+{y}^{2}+10$ nearest the plane $x+2y-z=0.$ Identify the point on the plane.

$\left(\frac{47}{24},\frac{47}{12},\frac{235}{24}\right)$

Find the point in the plane $2x-y+2z=16$ that is closest to the origin.

A company manufactures two types of athletic shoes: jogging shoes and cross-trainers. The total revenue from $x$ units of jogging shoes and $y$ units of cross-trainers is given by $R\left(x,y\right)=-5{x}^{2}-8{y}^{2}-2xy+42x+102y,$ where $x$ and $y$ are in thousands of units. Find the values of x and y to maximize the total revenue.

$x=3$ and $y=6$

A shipping company handles rectangular boxes provided the sum of the length, width, and height of the box does not exceed $96$ in. Find the dimensions of the box that meets this condition and has the largest volume.

Find the maximum volume of a cylindrical soda can such that the sum of its height and circumference is $120$ cm.

$V=\frac{64,000}{\pi }\approx 20,372$ cm 3

find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
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Abhi
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Abhi
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salma
Commplementary angles
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
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Kristine 2*2*2=8
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how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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Cied
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Porter
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Azam
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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