Gis constructing a triangular sail for a sailboat. aluminum supports will run along the horizontal base and vertical height of the sail, respectively. after making some measurements, calculo determines that the height of the sail cannot exceed 12 feet, or else it will not be able to fit under the bridges in the area. in addition, the total weight of the supports cannot exceed 80 pounds, or else the boat will sink. if aluminum weighs 5 pounds per foot, what is the area of the largest sail calculo can build? fully justify your answer, doing each step you learned in the class prep. in particular, you must show that the area you found is indeed the largest possible.

A (max) = 1/2 ( 8 * 8 )   = 32 ft²

Step-by-step explanation:

We have two constrains

a) The height of the sail ( support) cannot exceed 12 feet

b) The total weight of the supports cannot exceed 80 pounds

The weight of a foot of aluminum is 5 .

If we call:

x lenght of base support,  y  height of the vertical support  and the sail shape (is a triangle) we have

A = 1/2* ( x + y )

We know that (given a constant perimeter rectangle, the maximun area  is the square

A = x * y                P = 2x +2y         y = (P - 2x ) ÷ 2  

A = x * ( P - 2x ) ÷ 2  ⇒    A = (Px -2x²) ÷ 2

if we get derivative      A´(x) = (P-4x)/2     ⇒  (P-4x)/2   = 0

P = 4x and    x = P/4

Now we can look an square as two straight triangles joined by the diagonal and these two triangles are of area maximun.

Therefore in our case the sail must be of 8 x 8 feet (base and height)

A (max) = 1/2 ( 8 * 8 )   = 32 ft²   and we will keep the condition of weigh

8 + 8 = 16 feet   and the supports weights  16 * 5  = 80 pounds

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47

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step-by-step explanation: