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Karo-lina-s [1.5K]

3 years ago

8

Fred and Sue want to fence in a rectangular area in their yard for a vegetable garden. Suppose that they have 30 feet of fencing

. Sue tells Fred that she wants to put up the fence in such a way as to obtain a rectangular garden with the largest possible perimeter. Fred replies that the perimeter will be the same for any rectangular garden that they construct as long as they use all of the fencing. Is Fred correct? Explain.

2 answers:

AlekseyPX3 years ago

7 0

Answer:

Yes, Fed is correct

Explanation:

We only get 30 feet of fencing, it is already the largest possible perimeter that we can do if we use all of the fencing.

larisa [96]3 years ago

6 0

Answer:

a=35x−x^2

Step-by-step explanation:

Having

70 ft of fencing with a

width of x

feet and knowing the perimeter of a rectangle is

p=2w+2l

we can state the length of the garden as:

70=2x+2l

and solving for l

we know the length with be:

2l=70−2x or l=35−x

And then knowing the formula for the area of a rectangle is

a=w⋅l

we can write the equation as:

a=x⋅(35−x)a=35x−x^2

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A closed, rectangular-faced box with a square base is to be constructed using only 36 m2 of material. What should the height h a

kkurt [141]

Answer:

$b=h=\sqrt{6}$ m

Step-by-step explanation:

Let

Bas length of box=b

Height of box=h

Material used in constructing of box=36 square m

We have to find the height h and base length b of the box to maximize the volume of box.

Surface area of box= $2b^2+4bh$

$2b^2+4bh=36$

$b^2+2bh=18$

$2bh=18-b^2$

$h=\frac{18-b^2}{2b}$

Volume of box, V= $b^2h$

Substitute the values

$V=b^2\times \frac{18-b^2}{2b}$

$V=\frac{1}{2}(18b-b^3)$

Differentiate w. r.t b

$\frac{dV}{db}=\frac{1}{2}(18-3b^2)$

$\frac{dV}{db}=0$

$\frac{1}{2}(18-3b^2)=0$

$\implies 18-3b^2=0$

$\implies 3b^2=18$

$b^2=6$

$b=\pm \sqrt{6}$

$b=\sqrt{6}$

The negative value of b is not possible because length cannot be negative.

Again differentiate w.r.t b

$\frac{d^2V}{db^2}=-3b$

At $b=\sqrt{6}$

$\frac{d^2V}{db^2}=-3\sqrt{6}$

Hence, the volume of box is maximum at $b=\sqrt{6}$ .

$h=\frac{18-(\sqrt{6})^2}{2\sqrt{6}}$

$h=\frac{18-6}{2\sqrt{6}}$

$h=\frac{12}{2\sqrt{6}}$

$h=\sqrt{6}$

$b=h=\sqrt{6}$ m

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