3 Optimization Problems

[kiwi28]

I have 3 questions that I need to make sure I got right answers on for a test tomorrow. Here they are:

1. A banquet hall is to be built in the shape of a rectangular region with a semicircle on each end. The perimeter of the room is to be 750 feet. What dimensions will make the total are as large as possible?

Primary Eq: A=xy
Secondary Eq: 750=2x+Pi(y)

A=(750x-2x^2)/Pi

A'=[(750-4x)Pi]/Pi^2

x=187.5

y=375/Pi

My answer was 375/Pi x 187.5 ft.




2. Find the point on f(x)=(x+1)^2 closest to the point (4,3).

Primary Eq: d=[(x-4)^2+(y-3)^2]^(1/2)
Secondary Eq: y=(x+1)^2

d=(x^4+4x^3+x^2-16x+20)^(1/2)

d'=2x^3+6x^2+x-8

x=.9455

y=3.7850

My answer was (.9455,3.7850)




3. A retangle is inscribed between the parabolas y=6x^2 and y=20-x^2. What is the maximum area of such a rectangle?

Primary Eq: A=2xy

Secondary Eq: y=20-7x^2

A=40x-14x^3

A'=40-42x^2

x=+/-(20/21)^(1/2)

y=40/3

A=2xy=26.024

My answer was 26.024 units^2

Thank you for your time, and I hope my answers are correct.

 

[skeeter]

1. for a fixed perimeter, max area will be when the room is strictly circular.

2. x-value is ok ... y-value is incorrect. (point is on the curve y = (x+1)^2)

3. correct

 

[soroban]

Hello, kiwi28!

I got the same answers . . . Excellent work!

1. A banquet hall is to be built in the shape of a rectangular region with a semicircle on each end.
The perimeter of the room is to be 750 feet.
What dimensions will make the total are as large as possible?

                     x
              * - - - - - - *
          *   :             :   *
        *     :             :     *
       *      :             :      *
              :             :
      *       :             :       *
      *       :y           y:       *
      *       :             :       *
              :             :
       *      :             :      *
        *     :             :     *
          *   :             :   *
              * - - - - - - *
                     x

Primary Eq: \(\displaystyle \,A \:= \:xy\)
Secondary Eq: \(\displaystyle \,750\:=\:2x\,+\,\pi y\;\;\Rightarrow\;\;y \:=\:\frac{750\,-\,2x}{\pi}\)

\(\displaystyle A\:=\:xy \:=\:x\left(\frac{750-2x}{\pi}\right) \:=\:\frac{1}{\pi}(750x\,-\,2x^2)\)

Then: \(\displaystyle \,A'\:=\:\frac{1}{\pi}(750\,-\,4x) \:=\:0\;\;\Rightarrow\;\;\L x\:=\: 187.5\)

Then: \(\displaystyle \,y\:=\:\frac{750\,-\,2(187.5)}{\pi}\;\;\Rightarrow\;\;\L y \:=\:\frac{375}{\pi}\)

2. Find the point on \(\displaystyle f(x)\:=\x\,+\,1)^2\) closest to the point \(\displaystyle (4,3)\)

                    |
      *             | *
                    |           o(4,3)
       *            |*
        *           *
          *       * |
      - - - - * - - + - - - - - -
             -1     |

Primary Eq: \(\displaystyle \,d\:=\:\left[(x\,-\,4)^2\,+\,(y\,-\,3)^2\right]^{\frac{1}{2}}\;\;\Rightarrow\;\;D\:=\:d^2\:=\x\,-\,4)^2\,+\,(y\,-\,3)^2\)
Secondary Eq: \(\displaystyle \,y\:=\x\,+\,1)^2\)

Then: \(\displaystyle \\:=\x\,-\,4)^2\,+\,\left([x\,+\,1]^2\,-\,4\right)^2 \:=\:x^4\,+\,4x^3\,+\,x^2\,-\,16x\,+\,20\)

Hence: \(\displaystyle \,D'\:=\:4x^3\,+\,12x^2\,+\,2x\,-\,16 \:=\:0\)

This cubic has three real roots,
. . but the only one in Quadrant I is: \(\displaystyle \,\L x\,=\,0.9455,\;y\,=\,3.7850\)

3. A rectangle is inscribed between the parabolas \(\displaystyle y\:=\:6x^2\) and \(\displaystyle y\:=\:20\,-\,x^2\)
What is the maximum area of such a rectangle?

                      |
                     ***
         o       *    |    *       o
              *-------+-------*
           *  :       |       :  *
          o   :       |      y:   o
              :       |       :
         *  o :   x   |   x   : o  *
              o-------+-------o
                 o    |    o
      - * - - - - - -ooo- - - - - - * - -
                      |

Primary Eq: \(\displaystyle \,A\:=\:2xy\)

Secondary Eq: \(\displaystyle \,y\:=\20\,-\,x^2)\,-\,6x^2\:=\:20\,-\,7x^2\)

\(\displaystyle A\:=\:2x(20\,-\,7x^2)\:=\:40x\,-\,14x^3\)

\(\displaystyle A'\:=\:40\,-\,42x^2\:=\:0\;\;\Rightarrow\;\;x\:=\:\pm\sqrt{\frac{20}{21}}\;\;\Rightarrow\;\;y\:=\:\frac{40}{3}\)

Therefore: \(\displaystyle \,A \:=\:2xy \:=\:2\left(\sqrt{\frac{20}{21}}\right)\left(\frac{40}{3}\right)\;\;\Rightarrow\;\;\L A\:\approx\:26.024\)