# Let $x$ and $y$ be positive real numbers such that $4x + 9y = 60.$ Find the maximum value of $xy.$

Let $$x$$ and $$y$$ be positive real numbers such that $$4x + 9y = 60.$$ Find the maximum value of $$xy.$$

I think this problem has to do with the use of AM-GM. Can someone help me find the solution?

Yes, you are right. In this case AM-GM is the most elementary method.

$$AM \ge GM\\ \frac{4x+9y}{2} \ge \sqrt{4x\times 9y}\\ 25 \ge xy$$

Hence maximum value of $$xy$$ is 25.

• I think your answer is the best one ; $+1$ – J. W. Tanner May 15 '20 at 3:12

If $$4x+9y=60,$$ then $$y=\dfrac{60-4x}9$$, so $$xy=\dfrac{(60-4x)x}9=\dfrac{-4x^2+60x}9$$

$$=\dfrac{-(4x^2-60x+225)+225}{9}=\dfrac{225-(2x-15)^2}9\le\dfrac{225}9=25.$$

Different ways to do this problem. From a geometric stand point (I deliberately avoid a typical Calculus approach because you tagged Pre-Calculus), you can consider $$y=c/x$$ and $$4x+9y=60$$ in the first quadrant and you want these graphs to be tangential at a point. Subbing $$y=c/x$$ into the linear equation and multiplying through by $$x$$ results in $$4x^2-60x+9c=0$$ and using the Discriminant $$b^2-4ac=0$$ (You want ONE solution!) gives you $$c=25$$. From here you can find the point of tangency and all the rest.

Solving for $$x$$ gives us

$$x=15-\frac94 y$$

Then $$xy$$ is

$$f(y)=xy=y\left(15-\frac 94 y\right)=-\frac94 y^2+15y$$

This has a critical point at

$$0=f'(y)=-\frac92 y+15$$

$$\Rightarrow y=\frac{30}{9}$$

Since this is the only critical point and $$xy$$ is a quadratic with a negative leading coefficient, we may conclude that maximum occurs at this $$y$$. Thus, the maximum of $$xy$$ is

$$xy=\frac{30}{9}\left(15-\frac 94 \cdot \frac{30}{9}\right)=25$$

$$y=(60-4x)/9$$ $$xy=\frac{x(60-4x)}{9}$$ $$=\frac{4x(15-x)}{9}$$ $$=\frac{-4(x^2-15x)}{9}$$ $$=\frac{-4((x-7.5)^2-225/4)}{9}$$ $$=\frac{-4(x-7.5)^2+225)}{9}$$ max value = 225/9=25, which occurs for $$x=7.5,y=10/3.$$