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(1) parameter = 6x + 2h + 5x*2 = = 16x + 2h = 360, h = 180 - 8x cm
Let M be the middle point of QS, QM = 3x cm
It's easy to know that QRM is a right triagle, and RM = 4x cm (the sides are in the ratio of 3:4:5)
A = PT*PQ + (1/2)*QS*RM = 6x*(180 - 8x) + (1/2)*6x*4x = 1080x - 36x²
(2) A' = 1080 - 72x = 0, x= 15
(3) when x < 15, A' > 0; when x > 15, x < 0; the stationary value is a maximum
Let M be the middle point of QS, QM = 3x cm
It's easy to know that QRM is a right triagle, and RM = 4x cm (the sides are in the ratio of 3:4:5)
A = PT*PQ + (1/2)*QS*RM = 6x*(180 - 8x) + (1/2)*6x*4x = 1080x - 36x²
(2) A' = 1080 - 72x = 0, x= 15
(3) when x < 15, A' > 0; when x > 15, x < 0; the stationary value is a maximum
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