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My friend has $280 to spend on a fence for her rectangular garden. She wants to use cedar fencing which costs $13/foot on one side, and cheaper metal fencing which costs $7/foot for the other three sides.

What are the dimensions of the garden with the largest area she can enclose?

length for cedar side = ___foot

width for other side = ___foot

What is the largest area that can be enclosed?

area = ____foot-squared

If necessary, round your answers accurate to two decimal places.

Sagot :

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What are the dimensions of the garden with the largest area she can enclose?

The dimensions of the garden with the largest area she can enclose are length for cedar side = 7 foot and width for other side =  10 foot

Let L = length of the fence and W = width of the fence.

Also, let L be the length of cedar fencing.

Since we require one side of cedar fencing and the other 3 sides of cheaper metal fencing, and the cedar fencing costs $13/foot and the cheaper material costs $7/foot, the total cost of fencing is C = 13L + 7L + 2W × 7

= 13L + 7L + 14W

= 20L + 14W.

Since the total amount my friend wants to spend on fencing is $280, C = 280.

So, 20L + 14W = 280

10L + 7W = 140 (1)

Also, the area of the rectangular area is A = LW

From (1), L = (140 - 7W)/10

Substituting L into A, we have

A = LW

A = (140 - 7W)W/10

A = 14W - 0.7W²

To find the value of W that makes A maximum, we differentiate A with respect to W.

So, dA/dW = d(14W - 0.7W²)/dt

dA/dW = 14 - 1.4W

Equating it to zero, we have

14 - 1.4W = 0

14 = 1.4W

W = 14/1.4

W = 10 feet

To determine if this value maximizes A, we differentiate dA/dW with respect to W.

So, d²A/dW² = d(14 - 1.4W)/dW = -1.4

Since d²A/dW² = -1.4 < 0, W = 10 feet is a maximum point.

Since L = (140 - 7W)/10,

substituting W = 10 into the equation, we have

L = (140 - 7W)/10

L = (140 - 7 × 10)/10

L = (140 - 70)/10

L = 70/10

L = 7 feet

So, the dimensions of the garden with the largest area she can enclose are length for cedar side = 7 foot and width for other side =  10 foot

What is the largest area that can be enclosed?

The largest area that can be enclosed is area = 70 foot-squared

Since the area A = LW

Substituting the values of L and W into the e quation, we have

A = LW

A = 7 × 10

A = 70 ft²

So, the largest area that can be enclosed is area = 70 foot-squared

Learn more about dimensions of a rectangular garden here:

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