Charles planted a garden using 36 feet of fencing for the perimeter. The area of Charles's garden is 45 square feet. What are the dimensions of his garden?

Let P and A stand for the perimeter and area, respectively. Let L and W stand for the length and width, respectively.

We know ...

P = 2(L+W)

A = LW

The first equation can be solved for L, which can then be substituted into the second equation.

P/2 = L + W (divide both sides by 2)

P/2 - W = L (subtract W from both sides)

A = (P/2 - W)W (substitute for L)

A = WP/2 - W

W

W = (-(-P/2) ±√((-P/2)

= ((P/2) ±√((P

W = (P±√(P

These two solutions turn out to be L and W.

Example

P=36 ft, A=45 ft

W = (36±√(36

= (36±√576)/4

= (36±24)/4

= {3, 15}

The width is 3 ft, the length is 15 ft.

We know ...

P = 2(L+W)

A = LW

The first equation can be solved for L, which can then be substituted into the second equation.

P/2 = L + W (divide both sides by 2)

P/2 - W = L (subtract W from both sides)

A = (P/2 - W)W (substitute for L)

A = WP/2 - W

^{2}(use the distributive property)W

^{2}- WP/2 + A = 0 (add W^2 - WP/2 to both sides to put the quadratic into standard form)W = (-(-P/2) ±√((-P/2)

^{2}- 4(1)(A)))/(2(1))= ((P/2) ±√((P

^{2}-16A)/4)/2W = (P±√(P

^{2}-16A))/4These two solutions turn out to be L and W.

Example

P=36 ft, A=45 ft

^{2}W = (36±√(36

^{2}-16*45))/4= (36±√576)/4

= (36±24)/4

= {3, 15}

The width is 3 ft, the length is 15 ft.