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Wewaii [24]
3 years ago
12

The second side of a triangular deck is 5 feet longer than the shortest side and a third side that is 5 feet shorter than twice

the length of the shortest side. if the perimeter of the deck is 64 feet, what are the lengths of the three sides?
Mathematics
1 answer:
Luda [366]3 years ago
4 0
Let x be the length of the shortest side. second side is 5+x and the last is 2x-5. add these three to get the perimeter 64
x+5+x+2x-5=64
4x=64
x=16
sides are 16, 21, and 27
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\left[\begin{matrix} 0 & 9 \\ 9 & 0\end{matrix}\right]

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Since the function is continous and the region R is closed and bound (hence compact) the maximum and minimum must be attained on the boundaries of R. REcall that when -2\leq x \leq x and y=0 we have that F(x,0) = 0. So, we want to pay attention to the critical values over the circle, restricting that the values of y must be positive. To do so, consider the following function

H(x,y, \lambda) = 9xy - \lambda(x^2+y^2-4) which consists of the original function and a function that describes the restriction (the circle x^2+y^2=4), we want that the gradient of H is 0.

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From the first and second equation we get that

\lambda = \frac{9y}{2x} = \frac{9x}{2y}

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F(-\sqrt[]{2},\sqrt[]{2}) = 9\cdot -2 =-18.

Then, the point (\sqrt[]{2},\sqrt[]{2}) is a maximum and the point (-(\sqrt[]{2},\sqrt[]{2}) is a minimum.

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