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katovenus [111]

3 years ago

14

12. The point P lies on a line through the midpoint of AB and perpendicular to AB. If AP=15 in., what is the length PB? How do y

ou know?

1 answer:

Mekhanik [1.2K]3 years ago

8 0

Answer:

PB = 15

Step-by-step explanation:

A midpoint is a point in the center of a line. Since P is the midpoint of AB, the length of AP = PB. Therefore, if AP = 15, PB = 15.

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A landscape architect wished to enclose a rectangular garden on one side by a brick wall costing $20/ft and on the other three s

makvit [3.9K]

Answer:

The dimensions of the garden that minimize the cost is 9.018 feet(length) and 13.528 feet(width)

Step-by-step explanation:

Let the length of garden be x

Let the breadth of garden be y

Area of Rectangular garden = $Length \times Breadth = xy$

We are given that the area of the garden is 122 square feet

So, $xy=122$ ---A

A landscape architect wished to enclose a rectangular garden on one side by a brick wall costing $20/ft

So, cost of brick along length x = 20 x

On the other three sides by a metal fence costing $10/ft.

So, Other three side s = x+2y

So, cost of brick along the other three sides= 10(x+2y)

So, Total cost = 20x+10(x+2y)=20x+10x+20y=30x+20y

Total cost = 30x+20y

Substitute the value of y from A

Total cost = $30x+20(\frac{122}{x})$

Total cost = $\frac{2440}{x}+30x$

Now take the derivative to minimize the cost

$f(x)=\frac{2440}{x}+30x$

$f'(x)=-\frac{2440}{x^2}+30$

Equate it equal to 0

$0=-\frac{2440}{x^2}+30$

$\frac{2440}{x^2}=30$

$\sqrt{\frac{2440}{30}}=x$

$9.018 =x$

Now check whether it is minimum or not

take second derivative

$f'(x)=-\frac{2440}{x^2}+30$

$f''(x)=-(-2)\frac{2440}{x^3}$

Substitute the value of x

$f''(x)=-(-2)\frac{2440}{(9.018)^3}$

$f''(x)=6.6540$

Since it is positive ,So the x is minimum

Now find y

Substitute the value of x in A

$(9.018)y=122$

$y=\frac{122}{9.018}$

$y=13.528$

Hence the dimensions of the garden that minimize the cost is 9.018 feet(length) and 13.528 feet(width)

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