The dimensions of the garden with the largest area she can enclose are given by 24*14 square yards.
The area (A) of the rectangle with length L and Width W is given by,
A=L*W
The maximum of [tex]y=ax^2+bx[/tex] occurs at x = -b/2a
Let the length and width of the garden be L and W respectively.
Now let my friend use cedar fencing for one width and cheaper metal fencing for rest sides.
Rate of cedar fencing is $17/yard
Then she has to pay for cedar fencing = 17W
rate of cheaper metal fencing is $7/yard
Then she has to pay for metal fencing = 7W+7*2L = 7W+14L
Then according to condition,
17W+7W+14L = 672
24W+14L = 672
12W+7L = 336
7L = 336-12W
L = (336-12W)/7
Then the area of the rectangular garden is given by,
A = L*W
[tex]A=\frac{336-12W}{7}\times W\\A=-\frac{12}{7}W^2+48W[/tex]
So here a = -12/7 and b = 48
Then maximum of W occurs at,
W = -48/(2*(-12/7)) = (48*7)/(2*12) = 336/24 = 14
Then maximum width = 14 yards
then length (L) = (336-12*14)/7 = 168/7 = 24 yards
Hence the dimensions with the leargest area she can enclose is given by = 24 * 14 square yards.
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